Name: Solutions Which Deviate From Ideal Behaviour
Uploaded: 2019-07-19
Duration: 6 min 38 s
Description: Non-ideal solutions are the solutions that do not follow Raoult's law at all in their compositions. Let's discuss how these solutions deviate from ideal beha...

Question

A solution containing $30 g$ of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 kPa$ at $298 K$ . Further, $18 g$ of water is then added to the solution and the new vapour pressure becomes $2.9 kPa$ at $298 K$ . Calculate the vapour pressure at $298 K$ .

Accepted Answer

Moles of solute $(n_{2}) = \frac{Mass of solute}{Molar mass of solute} = \frac{30 g}{M_{2} {gmol}^{- 1}} = \frac{30}{M_{2}}$

Moles of solvent $(n_{1}) = \frac{Mass of solvent}{Molar mass of solvent} = \frac{90 g}{18 {gmol}^{- 1}} = 5$

$\frac{{P_{1}}^{0} - P_{1}}{{P_{1}}^{0}} = \frac{n_{2}}{n_{1} + n_{2}} \frac{{P_{1}}^{0} - 2.8}{{P_{1}}^{0}} = \frac{30 / M_{2}}{5 + 30 / M_{2}} 1 - \frac{2.8}{{P_{1}}^{0}} = \frac{30}{(5 M_{2} + 30)} 1 - \frac{30}{5 M_{2} + 30} = \frac{2.8}{{P_{1}}^{0}} 1 - \frac{6}{M_{2} + 6} = \frac{2.8}{{P_{1}}^{0}} \frac{M_{2} + 6 - 6}{M_{2} + 6} = \frac{2.8}{{P_{1}}^{0}} \frac{{P_{1}}^{0}}{2.8} = 1 + \frac{6}{M_{2}} - - - - (1)$

Further, if we add $18 g$ of water then the moles of water is $\frac{90 + 18}{18} = \frac{108}{18} = 6$

$P_{1} = 2.9 kPa \frac{{P_{1}}^{0} - P_{1}}{{P_{1}}^{0}} = \frac{30 M_{2}}{M_{2} (6 M_{2} + 30)} = \frac{5}{M_{2} + 5} 1 - \frac{2.9}{{P_{1}}^{0}} = \frac{5}{M_{2} + 5} \frac{{P_{1}}^{0}}{2.9} = \frac{M_{2} + 5}{M_{2}} = 1 + \frac{5}{M_{2}} \frac{{P_{1}}^{0}}{2.9} = 1 + \frac{5}{M_{2}} - - - (2)$

Divide equation $(1)$ by $(2)$ we get;

$\frac{2.9}{2.8} = \frac{1 + 6 / M_{2}}{1 + 5 / M_{2}} \frac{2.9}{2.8} = \frac{1 + 6 / M_{2}}{1 + 5 / M_{2}} 2.9 (1 + \frac{5}{M_{2}}) = 2.8 (1 + \frac{6}{M_{2}}) 2.9 + \frac{14.5}{M_{2}} = 2.8 + \frac{16.8}{M_{2}} M_{2} = 23$

The molar mass of solute $= 23 g {mol}^{- 1}$

Putting the value of molar mass in the equation $(1)$

$\frac{{P_{1}}^{0}}{2.8} = 1 + \frac{6}{23} = \frac{29}{23} {P_{1}}^{0} = 3.53 kPa$

The vapour pressure of water $= 3.53 kPa$

A solution containing $30 g$ of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 kPa$ at $298 K$ . Further, $18 g$ of water is then added to the solution and the new vapour pressure becomes $2.9 kPa$ at $298 K$ . Calculate the vapour pressure at $298 K$ .

Important Questions on Solutions

Learn and Prepare for any exam you want !

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate the vapour pressure at 298 K.

Important Questions on Solutions

Learn and Prepare for any exam you want !

A solution containing $30 g$ of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 kPa$ at $298 K$ . Further, $18 g$ of water is then added to the solution and the new vapour pressure becomes $2.9 kPa$ at $298 K$ . Calculate the vapour pressure at $298 K$ .