What is the resulting concentration (in molL) of aqueous hydrochloric acid if 500 mL of its gaseous form at 300 K and 100 kPa will be dissolved in 100 mL pure water Use 8.314 kPaL(Kmol)?

Answer 1

First, use the ideal gas law to find the number of moles of hydrochloric acid gas (HCl) in 500 mL at 300 K and 100 kPa. The formula is ( n = \frac{PV}{RT} ). Here, ( P = 100 , \text{kPa} ), ( V = 0.5 , \text{L} ), ( R = 8.314 , \text{kPa} \cdot \text{L} / (\text{K} \cdot \text{mol}) ), and ( T = 300 , \text{K} ).

Calculating gives ( n = \frac{100 \times 0.5}{8.314 \times 300} \approx 0.02 , \text{mol} ). When this amount is dissolved in 100 mL (0.1 L) of water, the concentration of HCl is ( \text{C} = \frac{n}{V} = \frac{0.02}{0.1} = 0.2 , \text{mol/L} ).