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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions

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 Answer

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$
$P_s=12.08kP_a$

 

answered Aug 5, 2014 by sreemathi.v
edited 1 day ago by sharmaaparna1
 
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