Answer

(i) $H_2 O$ :

The molecular mass of water, $H_2 O$

$=(2 \times$ Atomic mass of hydrogen $)+(1 \times$ Atomic mass of oxygen $)$

$=[2(1.0084)+1(16.00 u)]$

$=2.016 u+16.00 u$

$=18.016$

$=18.02 u$

(ii) $CO_2$

The molecular mass of carbon dioxide, $CO_2$

$=(1 \times$ Atomic mass of carbon $)+(2 \times$ Atomic mass of oxygen $)$

$=[1(12.011 u)+2(16.00 u)]$

$=12.011 u+32.00 u$

$=44.01 u$

(iii) $CH_4$

The molecular mass of methane, $CH_4$

$=(1 \times$ Atomic mass of carbon $)+(4 \times$ Atomic mass of hydrogen $)$

$=[1(12.011 u)+4(1.008 u)]$

$=12.011 u+4.032 u$

$=16.043 u$

Answer

The molecular formula of sodium sulphate is $Na_2 SO_4$.

Molar mass of $Na_2 SO_4=[(2 \times 23.0)+(32.066)+4(16.00)]$

$=142.066 g ~mol^{-1}$

Mass percent of an element

$ =\dfrac{\text{ Mass of that element in the compound }}{\text{ Molar mass of the compound }} \times 100 $

$\therefore$ Mass percent of sodium:

$=\dfrac{46.0 g}{142.066 g} \times 100$

$=32.379$ %

$=32.4 $ %

Mass percent of sulphur:

$=\dfrac{32.066 g}{142.066 g} \times 100$

$=22.57$ %

$=22.6 $ %

Mass percent of oxygen:

$=\dfrac{64.0 g}{142.066 g} \times 100 $

$=45.049 $ %

$=45.05 $ %

Answer

Percentage of iron by mass $=69.9 $ %

Molar mass of iron $=55.85$

Percentage of oxygen by mass $=30.1 $ %

Molar mass of oxygen $=16$

Now,

Relative mass of iron in iron oxide

$ =\dfrac{\text { Percentage of iron by mass }}{\text { At. mass of iron }}=\dfrac{69.9}{55.85}=1.25 $

Relative mass of oxygen in iron oxide

$ =\dfrac{\text { Percentage of oxygen by mass }}{\text { At. mass of oxygen }}=\dfrac{30.1}{16}=1.88 $

Now,

Simplest molar ratio of iron to oxygen $=1.25: 1.88=2: 3$

Hence the empirical formula of iron oxide is $ \mathrm{Fe}_2 \mathrm{O}_3$.

Answer

(i) 1 mole of carbon is burnt in air.

$ \mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 $

1 mole of C produces 1 mole of $ \mathrm{CO}_2$ which corresponds to $44 \mathrm{~g} {\text {~of } \mathrm{CO}_2}$ .

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

16 g of dioxygen corresponds to $\dfrac{16}{32}=0.5$ moles.

Here, dioxygen is the limiting reagent.

It will produce 0.5 moles of $ \mathrm{CO}_2$ which corresponds to $0.5 \times 44=22 \mathrm{~g} \mathrm{~of} \mathrm{~CO}_2$ .

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

16 g of dioxygen corresponds to $\dfrac{16}{32}=0.5 \mathrm{moles}$.

Here, dioxygen is the limiting reagent.

It will produce 0.5 moles of $ \mathrm{CO}_2$ which corresponds to $0.5 \times 44=22 \mathrm{~g} \mathrm{~of} \mathrm{~CO}_2$ .

Answer

$0.375 M$ aqueous solution of sodium acetate

$1000 ~ \text{mL} $ of solution containing $ 0.375$ moles of sodium acetate

$\therefore$ Number of moles of sodium acetate in $500 mL$

$=\dfrac{0.375}{1000} \times 500$

$=0.1875$ moles

Molar mass of sodium acetate $=82.0245 g$ mole $^{-1}$ (Given)

$\therefore$ Required mass of sodium acetate $=(82.0245 g ~ mol^{-1})(0.1875$ mole $)$

$=15.38 g$

Answer

Mass percent of nitric acid in the sample = 69 % [Given]

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass .

Molar mass of nitric acid $ (HNO_3) $

= $ {1 + 14 + 3 (16) } g~ mol^{-1} $

= 1 + 14 + 48

= $ 63 ~ g ~ mol^{-1} $

$ \therefore $ Number of moles in 69 g of $ (HNO_3) $

$ = \dfrac {69 ~ g}{63 ~ g ~ mol^{-1}} $

= 1.095 mol

Volume of 100 g of nitric acid solution

= $ \dfrac {Mass ~ of ~ solution}{Density ~ of ~ solution} $

= $ \dfrac {100 ~ g}{1.41 ~ g ~ mL^{-1}} $

= 70.92 mL $ \equiv 70.92 \times 10^{-3} L $

Concentration of nitric acid

= $ \dfrac {1.095~ mol}{70.92 \times 10^{-3} L } $

= $ 15.44 ~mol/L $

$ \therefore $ Concentration of nitric acid = $ 15.44~ mol/L $ .

Answer

1 mole of copper sulphate contains 1 mole of copper.

Molar mass of $CuSO_4=(63.5)+(32.00)+4(16.00)$

$=63.5+32.00+64.00$

$=159.5 g$

$159.5 g$ of $CuSO_4$ contains $63.5 g$ of copper.

$\Rightarrow 100 g$ of $CuSO_4$ will contain $\dfrac{63.5 \times 100 g}{159.5}$ of copper.

$\therefore$ Amount of copper that can be obtained from $100 g CuSO_4=\dfrac{63.5 \times 100}{159.5}$

$=39.81 g$

Answer

Mass percent of iron $(Fe)=69.9 \%$ (Given)

Mass percent of oxygen $(O)=30.1 \%$ (Given)

Number of moles of iron present in the oxide $=\dfrac{69.90}{55.85}$

$=1.25$

Number of moles of oxygen present in the oxide $=\dfrac{30.1}{16.0}$

$=1.88$

Ratio of iron to oxygen in the oxide,

$=1.25: 1.88$

$=\dfrac{1.25}{1.25}: \dfrac{1.88}{1.25}$

$=1: 1.5$

$=2: 3$

$\therefore$ The empirical formula of the oxide is $Fe_2 O_3$

Empirical formula mass of $Fe_2 O_3=[2(55.85)+3(16.00)] g$

Molar mass of $Fe_2 O_3=159.69 g$

$ \begin{aligned} \therefore n=\frac{\text{ Molar mass }}{\text{ Emprical formula mass }} & =\frac{159.69 g}{159.7 g} \\ & =0.999 \\ & =1(\text{ approx }) \end{aligned} $

Molecular formula of a compound is obtained by multiplying the empirical formula with $n$.

Thus, the empirical formula of the given oxide is $Fe_2 O_3$ and $n$ is 1 .

Hence, the molecular formula of the oxide is $Fe_2 O_3$.

Answer

The average atomic mass of chorine

$ =\left[\left(\text { fractional abundance of }{ }^{35} \mathrm{Cl}\right)\left(\text { molar mass of }{ }^{35} \mathrm{Cl}\right)+\left(\text { fractional abundance of }{ }^{37} \mathrm{Cl}\right)\left(\text { Molar mass of }{ }^{37} \mathrm{Cl}\right)\right] $

$ \left.\left.=\left[\left{\dfrac{(75.77)}{100}\right)(34.9689 ~ u)\right}+\left{\dfrac{(24.23)}{100}\right)(363.9659 ~u)\right}\right] $

$ =26.4959+8.9568\\=35.4527 \mathrm{u} $

$ \therefore \text { The average atomic mass of chlorine }=35.4527 \mathrm{u} $

Answer

(i) 1 mole of $C_2 H_6$ contains 2 moles of carbon atoms.

$\therefore$ Number of moles of carbon atoms in 3 moles of $C_2 H_6$

$=2 \times 3$

$=6$ mol

(ii) 1 mole of $C_2 H_6$ contains 6 moles of hydrogen atoms.

$\therefore$ Number of moles of carbon atoms in 3 moles of $C_2 H_6$

$=3 \times 6$

$=18$ mol

(iii) 1 mole of $C_2 H_6$ contains $6.023 \times 10^{23}$ molecules of ethane.

$\therefore$ Number of molecules in 3 moles of $C_2 H_6$

$=3 \times 6.023 \times 10^{23}\\=18.069 \times 10^{23}$

Answer

Molarity (M) of a solution is given by,

$=\dfrac{\text{ Number of moles of solute }}{\text{ Volume of solution in Litres }}$

$=\dfrac{\text{ Mass of sugar } / \text{ molar mass of sugar }}{2 L}$

$=\dfrac{20 g /[(12 \times 12)+(1 \times 22)+(11 \times 16)] g}{2 L}$

$=\dfrac{20 g / 342 g}{2 L}$

$=\dfrac{0.0585 mol}{2 L}$

$=0.02925 mol L^{{-1 }}$

$\therefore$ Molar concentration of sugar $=0.02925 ~mol ~L^{{-1 }}$

Answer

Molar mass of methanol $(CH_3 OH)=(1 \times 12)+(4 \times 1)+(1 \times 16)$

$=32 g ~mol^{- 1}$

$ \text { Molarity }(\mathrm{M})=\dfrac{\text { Number of moles }(\mathrm{n})}{\text { Volume of solution }\left(\mathrm{V}_{solution }\right)} $

$\text { Number of moles }(\mathrm{n})=\text { Molarity }(\mathrm{M})\times \text { Volume of solution }(\mathrm{V}_{ solution })$

$n=0.25 \mathrm{~mol} \mathrm{L} \times 2.5 \mathrm{~L}=0.625 \mathrm{~mol}$

$\text { Now, calculate the mass }(m) \text { of methanol: }$

$\text { Mass }(\mathrm{m})=\text { Number of moles }(\mathrm{n}) \times \text { Molecular mass }\left(M_\mathrm{m}\right) $

$m=0.625 \mathrm{~mol} \times 32 \mathrm{~g} / \mathrm{mol}=20 \mathrm{~g}$

$ 20 \mathrm{~g}=0.020 \mathrm{~kg} $

$ \begin{aligned} & \text { Density }(\rho)=\frac{\text { Mass }(\mathrm{m})}{\text { Volume }\left(\mathrm{V} _ \text {methanol }\right)} \ & \text { Volume }\left(\mathrm{V} _ \text {methanol }\right)=\frac{\text { Mass }(\mathrm{m})}{\operatorname{Density~}(\rho)} \ & V_{\text {methanol }}=\frac{0.020 \mathrm{~kg}}{0.793 \mathrm{~kg} / \mathrm{L}} \approx 0.02522 \mathrm{~L} \end{aligned} $

$ 0.02522 \mathrm{~L}=25.22 \mathrm{~mL} $

So, the volume of methanol needed is $\mathbf{2 5 . 2 2} \mathbf{~ m L}$.

Answer

Given:

We are given that the mass of air at sea level is $1034 \mathrm{~g} / \mathrm{cm}^2$. We need to convert this mass into $ \mathrm{kg} / \mathrm{m}^2$ to use SI units.

Convert Mass from $ \mathrm{g} / \mathrm{cm}^2$ to $ \mathrm{kg} / \mathrm{m}^2$

Convert grams to kilograms:

- 1 $ \mathrm{~g}=0.001 \mathrm{~kg}$

- Therefore, $1034 \mathrm{~g}=1034 \times 0.001 \mathrm{~kg}=1.034 \mathrm{~kg}$.

Convert $ \mathrm{cm}^{\mathbf{2}}$ to $ \mathrm{m}^{\mathbf{2}}$ :

- $ 1 \mathrm{~cm}=0.01 \mathrm{~m}$, so $1 \mathrm{~cm}^2=(0.01 \mathrm{~m})^2=0.0001 \mathrm{~m}^2$.

- Therefore, $1 \mathrm{~cm}^2=10,000 \mathrm{~m}^2$ (since $1 \mathrm{~m}^2=10,000 \mathrm{~cm}^2$ ).

- Thus, $1034 \mathrm{~g} / \mathrm{cm}^2=1.034 \mathrm{~kg} / 0.0001 \mathrm{~m}^2=1.034 \times 10,000 \mathrm{~kg} / \mathrm{m}^2=10340 \mathrm{~kg} / \mathrm{m}^2$.

Calculate the Force (Weight) of the Air:

The force (weight) can be calculated using the formula:

Force $=$ mass $\times g$

where $g$ (acceleration due to gravity) is approximately $9.81 \mathrm{~m} / \mathrm{s}^2$.

$ \text { Force }=10340 \mathrm{~kg} / \mathrm{m}^2 \times 9.81 \mathrm{~m} / \mathrm{s}^2 $

Calculate the Pressure:

Pressure is defined as force per unit area:

$ \text { Pressure }=\dfrac{\text { Force }}{\text { Area }} $

Since we are already working with force per unit area $\left(\mathrm{kg} / \mathrm{m}^2\right)$, the pressure will be in pascal ( Pa ).

Calculating the force:

$ \text { Force }=10340 \times 9.81=101325.4 \mathrm{~N} / \mathrm{m}^2 $

Thus, the pressure at sea level is approximately:

$ \text { Pressure } \approx 101325.4 \mathrm{~Pa} $

Expressing in Standard Form

To express this in standard form:

$ 101325.4 \mathrm{~Pa}=1.013254 \times 10^5 \mathrm{~Pa} $

The pressure at sea level is approximately $1.013 \times 10^5 \mathrm{~Pa}$.

Answer

SI Unit of Mass: The SI unit of mass is the kilogram, abbreviated as kg .

Definition of Kilogram: The kilogram is defined as the mass of the international prototype of the kilogram.

This prototype is a cylinder made of platinum-iridium, which is stored in an airtight jar.

The international prototype of the kilogram is kept at the International Bureau of Weights and Measures, which is located in France.

It is important to remember this definition as it is fundamental to the understanding of mass in the SI system.

Answer

Answer

Significant Figures: Significant figures are the digits in a number that contribute to its precision. This includes all non-zero digits, any zeros between significant digits, and any trailing zeros in the decimal portion.

When measuring a quantity, the number of significant figures indicates the precision of the measurement. For example, if a volume is measured as 15.6 mL , the three digits ( 1,5 , and 6 ) represent the precision of that measurement.

• Non-zero digits are always significant.

• Any zeros between significant digits are also significant.

• Leading zeros (zeros before the first non-zero digit) are not significant.

• Trailing zeros in a decimal number are significant. For example, in 10.50, all four digits are significant.

Examples:

• In the number 10.5 kg , there are three significant figures ( 1,0 , and 5 ).

• In the number 0.0045 , there are two significant figures ( 4 and 5 ), as the leading zeros are not counted.

• In the number 200, there is only one significant figure unless specified otherwise (like 200 .which would indicate three significant figures).

Using the correct number of significant figures is crucial in scientific measurements as it reflects the precision of the measurement and helps avoid misinterpretation of data.

Answer

(i)

Express the contamination level in percent by mass

Given:

Contamination level: 15 ppm (parts per million)

This means 15 parts of chloroform in 1,000,000 parts of water by mass.

Convert ppm to percent:

Percent by mass is calculated as parts per 100.

We know 1 ppm = 1 part per 1,000,000 parts.

Therefore, $15 \mathrm{ppm}=15$ parts per 1,000,000 parts.

Calculate percent by mass:

Percent by mass $=\left(\dfrac{15 \text { parts }}{1,000,000 \text { parts }}\right) \times 100$

Percent by mass $=\left(\dfrac{15}{1,000,000}\right) \times 100$

Percent by mass $=15 \times 10^{-6} \times 100$

Percent by mass $=15 \times 10^{-4} %$

So, the contamination level in percent by mass is $1.5 \times 10^{-3} %$.

(ii)

Determine the molality of chloroform in the water sample:

Given:

Contamination level: 15 ppm by mass.

This means 15 grams of chloroform in 1,000,000 grams of water.

Calculate the molar mass of chloroform ( $ \mathrm{CHCl}_3$ ):

Carbon (C): $12 \mathrm{~g} / \mathrm{mol}$

Hydrogen (H): $1 \mathrm{~g} / \mathrm{mol}$

Chlorine (CI): $35.5 \mathrm{~g} / \mathrm{mol}$

Therefore, molar mass of $ \mathrm{CHCl}_3$ :

Molar mass of $ \mathrm{CHCl}_3=12+1+(3 \times 35.5)=12$

$ +1+106.5=119.5 \mathrm{~g} / \mathrm{mol} $

Calculate the number of moles of chloroform:

Given mass of chloroform: 15 grams

Molar mass of chloroform: $119.5 \mathrm{~g} / \mathrm{mol}$

Number of moles of chloroform:

$\text { Number of moles } =\dfrac{\text { Given mass }}{\text { Molar mass }}$

$=\dfrac{15 \mathrm{~g}}{119.5 \mathrm{~g} / \mathrm{mol}} $

$=0.125 \text { moles }$

Calculate the molality:

Molality ( $m$ ) is defined as the number of moles of solute per kilogram of solvent.

Given mass of water: $1,000,000$ grams $=1,000 \mathrm{~kg}$

$ \text { Molality }= \dfrac{\text { Number of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}} $

$=\dfrac{0.125 \text { moles }}{1,000 \mathrm{~kg}}$

Molality $=0.125 \times 10^{-3} \mathrm{~mol} / \mathrm{kg}$

Molality $=1.25 \times 10^{-4} \mathrm{~mol} / \mathrm{kg}$

So, the molality of chloroform in the water sample is $1.25 \times 10^{-4} \mathrm{~mol} / \mathrm{kg}$.

Answer

(i) $0.0048 \text{can be written in scientific notation as} 4.8 \times 10^{-3}$

(ii) $234,000 \text{can be written in scientific notation as} 2.34 \times 10^{5}$

(iii) $8008 \text{can be written in scientific notation as} 8.008 \times 10^{3}$

(iv) $500.0 \text{can be written in scientific notation as} 5.000 \times 10^{2}$

(v) $6.0012 \text{can be written in scientific notation as} 6.0012$

Answer

(i) 0.0025 contains 2 significant figures.

(ii) 208 contains 3 significant figures.

(iii) 5005 contains 4 significant figures.

(iv) 126,00 contains 3 significant figures.

(v) 500.0 contains 4 significant figures.

(vi) 2.0034 contains 5 significant figures.

Answer

(i) The number 34.216 is rounded to three significant figures as 34.2 .

(ii) The number 10.4107 is rounded to three significant figures as 10.4 .

(iii) The number 0.04597 is rounded to three significant figures as 0.046 .

(iv) The number 2808 is rounded to three significant figures as 2810.

Answer

(a) When the mass of dinitrogen is 28 g , mass of dioxygen combined is $32,64,32$, and 80 g . The corresponding ratio is 1:2:1:5. It is a simple whole number ratio .

This illustrates the law of multiple proportions.

It states that, “if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

(b) (i) $1 \mathrm{~km}=1000 \mathrm{~m} \times 100 \mathrm{~cm} \times 10 \mathrm{~mm}$

$ \begin{aligned} & \therefore 1 \mathrm{~km}=10^6 \mathrm{~mm} \ & 1 \mathrm{~km}=1000 \mathrm{~m} \times \frac{1 \mathrm{pm}}{10^{-12} \mathrm{~m}} \ & \therefore 1 \mathrm{~km}=10^{15} \mathrm{pm} \end{aligned} $

Hence, $1 \mathrm{~km}=10^6 \mathrm{~mm}=10^{15} \mathrm{pm}$

(ii) $1 \mathrm{mg}=10^{-6} \mathrm{~kg}=10^6 \mathrm{ng}$

(iii) $1 \mathrm{ml}=10^{-3} \mathrm{l}=10^{-3} \mathrm{dm}^3$

Answer

Given :

Speed of light $=3 \times 10^8 \mathrm{~ms}^{-1}$

Times $=2.00 \mathrm{nsec}=2 \times 10^{-9} \mathrm{~s}$

As we know,

Distance covered $=$ Speed $\times$ time

The distance traveled is the product of speed and time.

Thus, in 2.00 nsec , the light will travel a distance of

$3.00 \times 10^8 \mathrm{~ms}^{-1} \times 2.00 \times 10^{-9} \mathrm{~s}$

$=0.600 \mathrm{~m}$

Answer

A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, $B$ is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of B will not be consumed. Hence, $A$ is the limiting reagent.

(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, $2.5 ~mol$ of B will combine with only $2.5 ~mol$ of A. As a result, $2.5 ~mol$ of A will be left as such. Hence, $B$ is the limiting reagent.

(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only $2.5 ~mol$ of B and the remaining $2.5 ~mol$ of B will be left as such. Hence, $A$ is the limiting reagent.

Answer

$ \underset{\substack{1 \mathrm{~mol} \=28.0 \mathrm{~g}}}{\mathrm{N_2}(\mathrm{~g})}+\underset{\substack{3 \mathrm{~mol} \=6.0 \mathrm{~g}}}{3 \mathrm{H}{2}(\mathrm{~g})} \rightarrow \underset{\substack{2 \mathrm{~mol} \=34.0 \mathrm{~g}}}{2 \mathrm{NH}{3}(\mathrm{~g})} $

(i)

28.0 g of $ \mathrm{N}_2$ require 6.0 g of $ \mathrm{H}_2$ to produce

$ =34.0 \mathrm{~g} \text { of } \mathrm{NH}_3 $

$2.00 \times 10^3 \mathrm{~g}$ of $ \mathrm{N}_2$ will produce

$ \begin{aligned} & =\frac{34}{28} \times 2.00 \times 10^3 \mathrm{~g} \text { of } \mathrm{NH}_3 \ & =2.43 \times 10^3 \mathrm{~g} \text { of } \mathrm{NH}_3 \ & =2430 \mathrm{~g} \mathrm{NH}_3 \end{aligned} $

(ii)

Yes, dihydrogen will remain unreacted to some extent

(iii)

Amount of hydrogen that remains unreacted.

28.0 g of $ \mathrm{N}_2$ require 6.0 g of $ \mathrm{H}_2$

$2.00 \mathrm{~g} \times 10^3 \mathrm{~g}$ of $ \mathrm{N}_2$ will require

$ =\frac{6.0}{28.0} \times 2.00 \times 10^3 \text { of } \mathrm{H}_2=428.5 \mathrm{~g} \text { of } \mathrm{H}_2 $

$\therefore$ Amount of hydrogen that remains unreacted

$ =\left[1.00 \times 10^3-428.5\right]_{\mathrm{g}} \ =571.5 \mathrm{~g} . $

Answer

Molar mass of $Na_2 CO_3=(2 \times 23)+12.00+(3 \times 16)$

$=106~g mol^{-1}$

Now, 1 mole of $Na_2 CO_3$ means $106 g$ of $Na_2 CO_3$.

$\therefore 0.5 mol$ $of$ $Na_2 CO_3=\dfrac{106 g}{1 mole} \times 0.5 mol~ Na_2 CO_3$

$=53 g Na_2 CO_3$

$\Rightarrow 0.50 M$ of $Na_2 CO_3=0.50~ mol / L~ Na_2 CO_3$

Hence, $0.50 ~mol$ of $Na_2 CO_3$ is present in $1 L$ of water or $53 g$ of $Na_2 CO_3$ is present in $1 L$ of water.

Answer

The balanced chemical equation for the reaction is,

$ 2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O} $

The volume of a gas is directly proportional to the number of moles.

Thus, 2 volumes of dihydrogen react with 1 volume of oxygen to form 2 volumes of water.

Thus, 10 volumes of dihydrogen react with 5 volume of oxygen to form 10 volumes of water.

Answer

(i) $28.7 pm$ :

$1 pm=10^{-12} m$

$ \therefore 28.7 pm = 28.7 \times 10^{-12} m $

$=2.87 \times 10^{{-11}} m$

(ii) $15.15 pm$ :

$1 pm=10^{-1} m$

$\therefore 15.15 pm=15.15 \times 10^{{-12}} m$

$=1.515 \times 10^{- 11} m$

(iii) $25365 mg$ :

$1 mg=10^{-3} g$

$25365 mg=2.5365 \times 10^{4} \times 10^{-3} g$

Since,

$1 g=10^{{-3}} kg$

$2.5365 \times 10^{1} g=2.5365 \times 10^{1} \times 10^{-3} kg$

$\therefore 25365 mg=2.5365 \times 10^{- 2} kg$

Answer

(i)

$1 \mathrm{~g} \mathrm{Au}=\frac{1}{197} \mathrm{~mol}$ atom of Au

$=\frac{1}{197} \times 6.022 \times 10^{23}$ atoms of Au

(ii)

$1 \mathrm{~g} \mathrm{Na}=\frac{1}{23} \mathrm{~mol}$ atom of Na

$=\frac{1}{23} \times 6.022 \times 10^{23}$ atom of Na

(iii)

$1 \mathrm{~g} \mathrm{Li}=\frac{1}{7} \mathrm{~mol}$ atom of Li

$=\frac{1}{7} \times 6.022 \times 10^{23}$ atom of Li.

(iv)

$1 \mathrm{~g} \mathrm{Cl}_2=\frac{1}{71}$ mol molecules of $ \mathrm{Cl}_2$

$=\frac{1}{71} \times 6.022 \times 10^{23}$ molecules of $ \mathrm{Cl}_2$

As one molecule of $Cl_2$ contains two atoms of $Cl$.

$=\frac{2}{71} \times 6.022 \times 10^{23}$ atoms of $ \mathrm{Cl}_2$

Hence, 1 g lithium has the largest number of atoms.

Answer

Mole fraction of $C_2 H_5 OH=\dfrac{\text{ Number of moles of } C_2 H_5 OH}{\text{ Number of moles of solution }}$

$0.040=\dfrac{n _{C_2 H_5 OH}}{n _{C_2 H_3 OH}+n _{H_2 O}}$

Number of moles present in $1 L$ water:

$n _{H_2 O}=\dfrac{1000 g}{18 g mol^{-1}}$

$n _{H_2 O}=55.55 mol$

Substituting the value of $n _{H_2 O}$ in equation (1),

$\dfrac{n _{C_2 H_5 OH}}{n _{C_2 H_3 OH}+55.55}=0.040$

$n _{C_2 H_3 OH}=0.040 n _{C_2 H_3 OH}+(0.040)(55.55)$

$0.96 n _{C_2 H_5 OH}=2.222 mol$

$n _{C_2 H_5 OH}=\dfrac{2.222}{0.96} mol$

$n _{C_2 H_3 OH}=2.314 mol$

$\therefore$ Molarity of solution $=\dfrac{2.314 mol}{1 L}$

$=2.314 M$

Answer

1 mole of carbon atoms $=6.023 \times 10^{23}$ atoms of carbon

$=12 g$ of carbon

$\therefore$ Mass of one ${ }^{12} C$ atom $=\dfrac{12 g}{6.022 \times 10^{23}}$

$=1.993 \times 10^{-23} g$

Answer

(i) The answer of the calculation $\dfrac{0.02856 \times 298.15 \times 0.112}{0.5785}$ will contain 3 significant figures as 0.112 contains 3 significant figures which is the lowest number of significant figures.

(ii) The answer to the calculation $5 \times 5.364$ will contain 4 significant figures as 5.364 contains 4 significant figures which are the lowest number of significant figures. Here, the exact figure 5 is not considered.

(iii) The answer to the calculation $0.0125+0.7864+0.0215$ will contain 4 significant figures as the result will be reported to 4 decimal places.

Answer

Molar mass of argon

$ \begin{aligned} & =[(35.96755 \times \frac{0.337}{100})+(37.96272 \times \frac{0.063}{100})+(39.9624 \times \frac{90.60}{100})] gmol^{-1} \\ & =[0.121+0.024+39.802] gmol^{-1} \\ & =39.947 gmol^{-1} \end{aligned} $

Answer

(i) 1 mole of $Ar=6.022 \times 10^{23}$ atoms of $Ar$

$\therefore 52 mol$ of $Ar=52 \times 6.022 \times 10^{23}$ atoms of $Ar$

$=3.131 \times 10^{25}$ atoms of $Ar$

(ii) 1 atom of $He=4 u$ of $He$

Or,

$4 u$ of $He=1$ atom of $He$

$1 u$ of $He=\frac{1}{4}$ atom of $He$

$52 u$ of $He = \frac{52}{4}$ atom of $He$

$=13$ atoms of $He$

(iii) $4 g$ of $He=6.022 \times 10^{23}$ atoms of $He$

$\therefore 52 g$ of $He=\dfrac{6.022 \times 10^{23} \times 52}{4}$ atoms of $He$

$=7.8286 \times 10^{24}$ atoms of $He$

Answer

(i)

1 mole $(44 g)$ of $CO_2$ contains $12 g$ of carbon.

$ 3.38 g \text{ of } CO_2 \text{ will contain carbon } = \dfrac{12g}{44g} \times 3.38g $

$=0.9217 g$

$18 g$ of water contains $2 g$ of hydrogen.

$\therefore 0.690 g$ of water will contain hydrogen

$ =\dfrac{2 g}{18 g} \times 0.690 $

$=0.0767 g$

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

$=0.9217 g+0.0767 g$

$=0.9984 g$

$\therefore$ Percent of $C$ in the compound

$ =\dfrac{0.9217 g}{0.9984 g} \times 100 $

$=92.32 %$

Percent of $H$ in the compound

$ =\dfrac{0.0767 g}{0.9984 g} \times 100 $

$=7.68 %$

Moles of carbon in the compound $=\dfrac{92.32}{12.00}$ $=7.69$

Moles of hydrogen in the compound $=\dfrac{7.68}{1}$

$=7.68$

$\therefore$ Ratio of carbon to hydrogen in the compound $=7.69: 7.68$

$=1: 1$

Hence, the empirical formula of the gas is $CH$.

(ii)

Given,

Weight of 10.0L of the gas (at S.T.P) $=11.6 g$

$\therefore$ Weight of $22.4 L$ of gas at STP

$ =\dfrac{11.6 g}{10.0 L} \times 22.4 L $

$=25.984 g$

$= 26g $

Hence, the molar mass of the gas is $26 g$.

(iii)

Empirical formula mass of $CH=12+1=13 g$

$ \begin{aligned} n & =\dfrac{\text{ Molar mass of gas }}{\text{ Empirical formula mass of gas }} \\ & =\dfrac{26 g}{13 g} \\ n & =2 \\ \therefore & \text{ Molecular formula of gas }=(CH) _{n} \\ = & C_2 H_2 \end{aligned} $

Answer

$0.75 M$ of $HCl \quad 0.75 mol$ of $HCl$ are present in $1 L$ of water

$ = [(0.75 \text{mol}) \times (36.5 \text{g mol}^{-1})] HCl$ is present in $1 L$ of water

$ = 27.375 g$ of $HCl$ is present in $1 L$ of water

Thus, $1000 mL$ of solution contains $27.375 g$ of $HCl$.

$\therefore$ Amount of $HCl$ present in $25 mL$ of solution

$=\dfrac{27.375 g}{1000 mL} \times 25 mL$

$=0.6844 g$

From the given chemical equation,

$ CaCO _{3(s)}+2 HCl _{(a q)} \longrightarrow CaCl _{2(a q)}+CO _{2(g)}+H_2 O _{(l)} $

$2 mol$ of $HCl(2 \times 36.5=71 g)$ react with $1 mol$ of $CaCO_3(100 g)$.

$\therefore$ Amount of $CaCO_3$ that will react with $0.6844 g$

$ =\dfrac{100}{71} \times 0.6844 g $

$=0.9639 g$

Answer

According to the given equation:

$ \mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} $

1 mole of $ \mathrm{MnO}_2$ reacts with 4 mole of HCl

or $87 \mathrm{g} \mathrm{MnO}_2$ reacts with 146 g HCl

$\therefore 5 \mathrm{g} \mathrm{MnO}_2$ Will react with $\dfrac{146}{87} \times 5 \mathrm{g} \mathrm{HCl}=8.39 \mathrm{HCl}$