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Byju's Answer

Standard XII

Chemistry

Relative Lowering of Vapour Pressure

The total vap...

Question

The total vapour pressure of a 4 mole % solution of ammonia in water at 293 K is 50.0 torr; the vapour pressure of pure water is 17.0 torr at this temperature. Applying Henry's and Raoult's laws, the total vapour pressure for a 5 mole % solution is (in torr).

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Solution

The given data is
$P_{w a t e r}^{o}$ =17.0 torr;
$P_{t o t a l}$ (4 mole % solution) = $P_{N H_{3}} + P_{w a t e r}$ =50.0 torr
$x_{N H_{3}}$ = 0.04 and ; $x_{w a t e r}$ = 0.96
Now according to Raoult's law;
$P_{w a t e r}$ = $x_{w a t e r}$ $P_{w a t e r}^{0}$
= 0.96 $\times$ 17.0 torr = 16.32 torr, $P_{a m m o n i a} = 33.68$ torr
Now Henry's law constant for ammonia is
$K_{H}$ ( $N H_{3}$ )= $\frac{P_{N H_{3}}}{x_{N H_{3}}}$ =842 torr
Hence, for 5 mole % solution, we have $P_{N H 3}$ = $K_{H}$ $(N H_{3})$ $x_{N H_{3}}$
= (842 torr) (0.05) = 42.1 torr
$P_{w a t e r}$ = $x_{w a t e r}$ $P_{w a t e r}^{0}$ = (17 torr) (0.95) =16.15 torr
Thus, $P_{t o t a l}$ (5 mole % solution) = $P_{N H 3}$ + $P_{w a t e r}$ = 42.1 + 16.15= 58.25 torr

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