A closed end manometer is filled with Hg as show in the following diagram at 27°C and volume of the bulb is 150 mL. At this temperature, the vapour pressure of water is 28 mm of Hg. If the bulb contains 0.001 mol of $O_{2} (g)$ then find the volume of liquid water. (Given that $R = 0.082 L . atm {mol}^{- 1} K^{- 1}$ ). Neglect volume of U-tube of manometer.

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34 mL

10 mL

27 mL

Data insufficient

answer is C.

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Detailed Solution

Pressure on oxygen,

$\begin{array}{l} P_{O_{2}} = 180 - 28 = 152 mm \\ V_{O_{2}} \times \frac{152}{760} = 0 .001 \times 0 .082 \times 300 \\ V_{O_{2}} = 123 mL \\ ∴ V_{H_{2} O} = 150 - 123 = 27 mL \end{array}$

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