States of Matter - Result Question 2-2
2. At $100^{\circ} \mathrm{C}$ and 1 atm if the density of the liquid water is $1.0 \mathrm{~g} \mathrm{~cm}^{-3}$ and that of water vapour is $0.0006 \mathrm{~g} \mathrm{~cm}^{-3}$, then the volume occupied by water molecules in 1 L of steam at this temperature is
(2000, 1M)
(a) $6 \mathrm{~cm}^3$
(b) $60 \mathrm{~cm}^3$
(c) $0.6 \mathrm{~cm}^3$
(d) $0.06 \mathrm{~cm}^3$
Show Answer
Answer:
Correct Answer: 2. (c)
Solution:
Let us consider, 1.0 L of liquid water is converted into steam.
Volume of $\mathrm{H}_2 \mathrm{O}(l)=1 \mathrm{~L}$, mass $=1000 \mathrm{~g}$
$\Rightarrow \quad$ Volume of 1000 g steam $=\dfrac{1000}{0.0006} \mathrm{~cm}^3$
$\because$ Volume of molecules in $\dfrac{1000}{0.0006} \mathrm{~cm}^3$ steam $=1000 \mathrm{~cm}^3$
$\therefore$ Volume of molecules in
$ 1000 \mathrm{~cm}^3 \text { steam }=\dfrac{1000}{1000} \times 0.0006 \times 1000=0.60 \mathrm{~cm}^3 $
BETA
