States of Matter 2 Question 2
2. At $100^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ 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 $1 \mathrm{~L}$ 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) $0.06 \mathrm{~cm}^{3}$
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Answer:
Correct Answer: 2. (c)
Solution:
- Let us consider, $1.0 \mathrm{~L}$ 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 $1000 \mathrm{~g}$ steam $=\frac{1000}{0.0006} \mathrm{~cm}^{3}$
$\because$ Volume of molecules in $\frac{1000}{0.0006} \mathrm{~cm}^{3}$ steam $=1000 \mathrm{~cm}^{3}$
$\therefore$ Volume of molecules in
$$ 1000 \mathrm{~cm}^{3} \text { steam }=\frac{1000}{1000} \times 0.0006 \times 1000=0.60 \mathrm{~cm}^{3} $$
