The Henry's law constant for helium gas in water at 30 °C is 3.70 × $10 ^ { - 4 }$ mol L^-1 bar^-1.When the partial pressure of helium above a sample of water is 0.450 bar,the concentration of helium in the water is ________ mol L^-1. A)1.22 × 10³ B)8.22 × 10^-4 C)1.67 × 10^-4 D)0.45 E)3.70 × 10^-4

Question

Quizplus · Accepted Answer

Use the formula for Henry's law:
concentration = partial pressure / Henry's law constant
concentration = 0.450 bar / 3.70 × 10^-4 mol L^-1 bar^-1 
concentration = 1.22 × 10^-3 mol L^-1 
But the units in the answer choices are not in mol L^-1, they are in mol. To convert, we need the volume of the water. This information is not given in the question, so we cannot calculate an exact number. Instead, we can use the fact that the answer choices are very different in magnitude and choose the best one based on that. 
Choice C is the most reasonable because it is of similar magnitude to the Henry's law constant (10^-4 mol), whereas the other choices are either too small (10^-13 mol) or too large (0.45 mol).

The Henry's Law Constant for Helium Gas in Water at 30