Question

A sample of hydrogen gas is collected over water at 140 C. The pressure of the resultant mixture 113.0 kPa. What is the pressure that is exerted by the dry hydrogen alone? The vapor pressure of water at 140 C is 11.9 mmHg.

Answer 1

To find the pressure of the dry hydrogen gas alone, we need to subtract the pressure exerted by the water vapor from the total pressure of the mixture.

First, we need to convert the vapor pressure of water from mmHg to kPa:

11.9 mmHg x (1 kPa / 7.5006 mmHg) = 0.0159 kPa

Next, we need to subtract this value from the total pressure of the mixture:

113.0 kPa - 0.0159 kPa = 112.9841 kPa

Therefore, the pressure exerted by the dry hydrogen gas alone is 112.9841 kPa.

Step-by-step explanation: