Question

The vapour pressure of water is $12.3kPa$ at $300K$. Calculate the vapour pressure $1$ molal solution of a non-volatile solute in it.

• A. $12.08kPa$
• B. $1.208Ppa$
• C. $2.4kPa$
• D. $0.4kPa$
  

Answer 1

The correct option is A $12.08kPa$

$1$ molal solution means 1 mol of the solute is present in $1000g$ of the solvent (water).

The molar mass of water $=18g$ $mo{l}^{-1}$

Number of moles present in $1000$ g of water $=1000/18$

$=55.56mol$

Therefore, the mole fraction of the solute in the solution is

$x_2=1/(1+55.56)$

$=0.0177$

It is given that,

Vapour pressure of water, $P_1^0$

$=12.3kPa$

Applying the relation, $(P_1^0-P_1)/P_1^0=X_2$

$=(12.3-P_1)/12.3$

$=0.0177$

$=12.3-P_1$

$=0.2177$

$=P_1$

$=12.0823$

$=12.08kPa$