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1.00 g of water is introduced into a 5.00 L evacuated flask at 50° C.
Vapour Pressure at 50° C
$H_{2} O$ 92.5 mmHg
What mass of water is present as liquid when equilibrium is established?

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0.59 g

0.083 g

0.41 g

answer is C.

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Detailed Solution

$n_{H_{2} O (g)} = \frac{\frac{92.5}{760} \times 5}{0.0821 \times 323} = 0.023 = 0.413 g$

$∴ w_{H_{2} O (l)} = 1 - 0.413 = 0.587 g$

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