The vapour pressure of water at $20^{o} C$ is $17.54 m m H g$ . Then the vapour pressure of the water in the apparatus is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (temp. constant) is:

5.77 m m H g

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16 m m H g

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35.08 m m H g

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between

8.77

and

15.54 m m H g

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Solution

The correct option is B $16 m m H g$
$We know that,$
$Mole fraction of sucrose: Moles of sucrose Moles of sucrose+ Moles of water$
So,
$X_{s u c r o s e} = \frac{\frac{200 g}{342.13 g m o l^{- 1}}}{\frac{200 g}{342.13 g m o l^{- 1}} + \frac{112.3 g}{18.01 g m o l^{- 1}}} = 0.0857$

Mole fraction of water:
$X_{w a t e r} = \frac{\frac{112.3 g}{18.01 g m o l^{- 1}}}{\frac{112.3 g}{18.01 g m o l^{- 1}} + \frac{200 g}{342.13 g m o l^{- 1}}} = 0.914$

Note that by definition, $x_{s u c r o s e} + x_{w a t e r} = 1$
Because sucrose is involatile, the vapour pressure of the solution is proportional to the mole fraction of water:
$P_{s o l u t i o n} = X_{w a t e r} + 17.54 m m$ of $H g$
$= 0.914 \times 17.54 m m$ of $H g = 16 m m$ of $H g$
Conclusion: hence the correct answer is not mentioned in the above options.

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The vapour pressure of water at 20 degree celsius is 17.54 mmHg. What will be the vapour pressure of the water in the apparatus shown after the piston is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (assume temperature constant).

A) 8.77 mmHg

B)17.54mmHg

C)35.08 mmHg

D) between 8.77 and 17.54mmHg

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