Some Basic Concepts of Chemistry - Result Question 5
5. 100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of $\mathrm{CaCO}_3$ is (molar mass of calcium bicarbonate is $162 \mathrm{~g} \mathrm{~mol}^{-1}$ and magnesium bicarbonate is $146 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(2019 Main, 8 April I)
(a) $5,000 \mathrm{ppm}$
(b) $1,000 \mathrm{ppm}$
(c) 100 ppm
(d) $10,000 \mathrm{ppm}$
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Answer:
Correct Answer: 5. (d)
Solution: Given, $W_{\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2}=0.81 \mathrm{~g}$
$$ \begin{aligned} & W_{\mathrm{Mg}_{\left(\mathrm{HCO}_3\right)2}}=0.73 \mathrm{~g} \\ & M{\mathrm{Ca}\left(\mathrm{HCO}_3\right)2}=162 \mathrm{~g} \mathrm{~mol}^{-1} \text {, } \\ & M{\mathrm{Mg}\left(\mathrm{HCO}_3\right)_2}=146 \mathrm{~mol}^{-1} \\ & \end{aligned} $$
$$ V_{\mathrm{H}_2 \mathrm{O}}=100 \mathrm{~mL} $$
Now, $\quad n_{\text {eq }}\left(\mathrm{CaCO}3\right)=n{\text {eq }}\left[\mathrm{Ca}\left(\mathrm{HCO}_3\right)2\right]+n{\text {eq }}\left[\mathrm{Mg}\left(\mathrm{HCO}_3\right)_2\right]$
$$ \begin{aligned} \dfrac{W}{100} \times 2 & =\dfrac{0.81}{162} \times 2+\dfrac{0.73}{146} \times 2 \\ \therefore \quad \dfrac{W}{100} & =0.005+0.005 \\ W & =0.01 \times 100=1 \end{aligned} $$
Thus, hardness of water sample $=\dfrac{1}{100} \times 10^6=10,000 \mathrm{ppm}$
BETA
