If 100 mL of HCl gas at 300 K and 100 kPa dissolve in 20 mL of pure water, what is the concentration?

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Answer 1 · 2014-08-10T18:20:57

The concentration of the HCl is 0.20 mol/L.

There are two steps to this calculation.

Step 1.

$PV = nRT$

$n = (PV)/(RT) = (100"kPa" × 0.100"L")/(8.314"kPa·L·K⁻¹mol⁻¹" × 300"K")$ = 4.01 × 10⁻³ mol

Step 2.

Molarity = #"moles of solute"/"litres of solution" = (4.01 × 10⁻³"mol")/(0.020"L")# = 0.20 mol/L

The concentration of the HCl is 0.20 mol/L.