Question

At $100^\circ C$ and $1atm$ , if the density of liquid water is $1.0gcm^{- 3}$ and that of water vapour is $0.0006\text{gc}\text{m}^{- 3}$ , then the volume occupied by water molecules in 1 litre of steam at that temperature

• A. $6cm^{3}$
• B. $60cm^{3}$
• C. $0.6cm^{3}$
• D. $0.06cm^{3}$

Answer 1

Answer: (C)

$\text{1L}=1000 \, \text{mL}=1000 \, \text{c}\text{m}^{3}$
$\text{mass}=\text{density}\times \text{volume}$
$=\left(0.0006 g cm ^{-3}\right) \times\left(1000 cm ^{3}\right)=0.6 g$
As 18 g of water $=\text{18 c}\text{m}^{3}$
So 0.6 g of water $=0.\text{6 c}\text{m}^{3}$
Hence actual volume occupied by molecules $=0.\text{6 c}\text{m}^{3}$