Question

At ${100}^o$C and $1$ atm, if the density of the liquid water is $1.0$ g/cc and that of water vapour is $0.00006$g/cc, then the volume occupied by water molecule in $1$L steam at this temperature is:

• A. $0.06$ cc
• B. $60$ cc
• C. $0.6$ cc
• D. none of the above

Answer 1

The correct option is A $0.06$ cc

We consider,1.0 L of liquid water is converted into steam volume of $H_{2}O$(l)= 1l,mass = 1000 g

Volume of 1000 g steam = $1000/0.00006c{m}^3$

Volume of molecules in $1000/0.00006c{m}^3$ steam = $1000c{m}^3$

volume of molecule in $1000c{m}^3$ steam = $\frac{1000\times 0.00006\times 1000}{1000}$$=0.06c{m}^3$

So option A is correct.