$50 c m^{3}$ of hydrogen is collected over water at $17^{\circ} C$ and 750 Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at $17^{\circ} C$ is 14 mm Hg.

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Solution

$\frac{P V}{T} = \frac{P_{1} V_{1}}{T_{1}}$
Here $P = 750 - 14 = 736 m m, V = 50 c m^{3}, T = 17 + 273 = 290 K, P_{1} = 760 m m, T_{1} = 273 K$
then, $V_{1} = \frac{736 \times 50 \times 273}{290 \times 760} = 45.58 c m^{3}$

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50cm3 of hydrogen is collected over water at 17∘C and 750 Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at 17∘C is 14 mm Hg.

PVT=P1V1T1Here P=750−14=736mm, V=50cm3, T=17+273=290K, P1=760mm, T1=273Kthen, V1=736×50×273290×760=45.58cm3

$50 c m^{3}$ of hydrogen is collected over water at $17^{\circ} C$ and 750 Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at $17^{\circ} C$ is 14 mm Hg.

$\frac{P V}{T} = \frac{P_{1} V_{1}}{T_{1}}$
Here $P = 750 - 14 = 736 m m, V = 50 c m^{3}, T = 17 + 273 = 290 K, P_{1} = 760 m m, T_{1} = 273 K$
then, $V_{1} = \frac{736 \times 50 \times 273}{290 \times 760} = 45.58 c m^{3}$