50 cm³ of gas was collected over water at 10°C and 765 mmHg. Calculate the volume of the gas at s.t.p. if the saturated vapour pressure of water at 10°C is 5 mmHg.

Correct answer: 48.23 cm³ (Option D)

Reasoning:

• Dry gas pressure at 10°C: Subtract water vapour pressure: \(P_{\text{dry}} = 765 - 5 = 760\ \text{mmHg}\).
• At s.t.p., \(P_2 = 760\ \text{mmHg}\) and \(T_2 = 273\ \text{K}\). Initial temperature \(T_1 = 273 + 10 = 283\ \text{K}\).
• Use the combined gas law for the dry gas: \(\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}\). Since \(P_1 = P_2\), then \(V_2 = V_1 \times \dfrac{T_2}{T_1} = 50 \times \dfrac{273}{283} \approx \mathbf{48.23\ \text{cm}^3}\).

Why others are wrong:

• 48.55 cm³ (Option C): Comes from not subtracting the water vapour pressure (using 765 mmHg instead of 760 mmHg).
• 48.87 cm³ and 49.19 cm³: Result from assorted rounding or using incorrect temperature/pressure corrections.