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

$\begin{array}{1 1}12.08kPa\\10.08kPa\\15.08kPa\\11.08kPa\end{array}$

How 55.5 comes in place of moles of solvent

1 molal solution means 1 mol of the solute in 1kg of the solvent(water)
Number of moles of water =1000 g / (18g/mol  )  =55.56 moles
Mole fraction of solute =$\large\frac{1}{1+55.5}$
$\Rightarrow 0.0177$
Now,$\large\frac{P_0-P_s}{P_0}=$$x_2 (ie) \large\frac{12.3-P_s}{12.3}$$=0.0177$
$P_s=12.08kP_a$

edited 1 day ago