A sample of hydrogen gas was collected over water at 21°C and 685 mmHg. The volume of the container was 7.80 L. Calculate the mass of H2(g) collected. (Vapor pressure of water = 18.6 mmHg at 21°C.)
A. 435 g
B. 7.14 g
C. 0.589 g
D. 0.572 g
E. 0.283 g
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Answer:
The total pressure of the gas is the sum of the partial pressure of the hydrogen gas and the vapor pressure of the water. Thus, the partial pressure of the hydrogen gas is 685 mmHg - 18.6 mmHg = 666.4 mmHg.
We can use the ideal gas law, which states that PV = nRT, to calculate the number of moles of hydrogen gas present in the container. Here, P is the partial pressure of the hydrogen gas, V is the volume of the container, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in kelvins.
Rearranging the ideal gas law to solve for n, we get n = PV/RT. Plugging in the values and using the ideal gas constant R = 0.0821 L-atm/mol-K, we get n = (666.4 mmHg * 7.80 L) / (0.0821 L-atm/mol-K * 294.15 K) = 0.0185 mol.
Since the molar mass of hydrogen gas is 2.016 g/mol, the mass of the hydrogen gas is 0.0185 mol * 2.016 g/mol = 0.037 g. Since the answer choices are given in grams, we must convert this value to grams. Therefore, the mass of the hydrogen gas is 0.037 g * 1000 mg/g = 37 mg.
Thus, the correct answer is B.7.14 G