If 200 cm³ of carbon (IV) oxide was collected over water at 18°C and 700 mmHg, calculate the volume of the dried gas at S.T.P. (Saturated vapour pressure of water at 18°C = 760 mmHg, Standard temperature = 273K)

The total pressure is the sum of the partial pressure of the dry gas and the water vapor pressure.

$$P_{dry} = P_{total} - P_{H_2O}$$ $$P_{dry} = 700 \text{ mmHg} - 15.5 \text{ mmHg}$$ $$P_{dry} = 684.5 \text{ mmHg}$$

The combined gas law is given by:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

Where:
$P_1 = 684.5 \text{ mmHg}$
$V_1 = 200 \text{ cm}^3$
$T_1 = 291 \text{ K}$
$P_2 = 760 \text{ mmHg}$ (Standard Pressure)
$T_2 = 273 \text{ K}$ (Standard Temperature)

Rearranging for $V_2$:

$$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$$ $$V_2 = \frac{684.5 \text{ mmHg} \times 200 \text{ cm}^3 \times 273 \text{ K}}{760 \text{ mmHg} \times 291 \text{ K}}$$ $$V_2 = \frac{37350300}{221160} \text{ cm}^3$$ $$V_2 \approx 168.90 \text{ cm}^3$$

The volume of the dried gas at S.T.P. is $\text{168.90 cm}^3$.