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JAMB Chemistry 2012 Paper

Question : 3

Total: 45

50 cm

_3

of gas was collected over water at 10ºC and 765 mm Hg. Calculate the volume of the gas at s.t.p. if the saturated vapour pressure of water at 10ºC is 5mm Hg

49.19 cm ³
48.87 cm ³
48.55 cm ³
48.23 cm ³

Validate

Solution:

Concept:

When gas is collected over water, the total pressure includes water vapour pressure. Remove it to get the dry gas pressure, then use the combined gas law to find the volume at s.t.p.

Explanation:

Dry gas pressure at the initial condition is:

P_1 = 765\ \mathrm{mm\,Hg} - 5\ \mathrm{mm\,Hg} = 760\ \mathrm{mm\,Hg}

Initial temperature is:

T_1 = 10^\circ\mathrm{C} + 273 = 283\ \mathrm{K}

At s.t.p., the conditions are:

P_2 = 760\ \mathrm{mm\,Hg}

and

T_2 = 273\ \mathrm{K}

Initial volume is

V_1 = 50\ \mathrm{cm}^3

Use the general gas equation:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Rearrange for

V_2

V_2 = \frac{P_1V_1T_2}{P_2T_1}

Since

P_1 = P_2 = 760\ \mathrm{mm\,Hg}

, they cancel out:

V_2 = \frac{50 \times 273}{283} = 48.23\ \mathrm{cm}^3

This matches the correct option.

Answer:

48.23\ \mathrm{cm}^3

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