If 500 mL of HCl gas at 300 K and 100 kPa dissolve in 100 mL of pure water, what is the concentration?

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If 500 mL of HCl gas at 300 K and 100 kPa dissolve in 100 mL of pure water, what is the concentration?

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Answer 1 · 2014-09-26T20:34:53

The concentration of HCl is 0.200 mol/L.

You use the Ideal Gas Law to find the number of moles, and then you calculate the molarity.

$P V = n R T$ $PV = nRT$

$n = \frac{P V}{R T} = \frac{100 kPa \times 0.500 L}{{8.314 kPa\cdotL\cdotK}^{- 1} {mol}^{- 1} \times 300 K}$ $n = (PV)/(RT) = "100 kPa × 0.500 L"/("8.314 kPa·L·K"^-1"mol"^-1 × "300 K")$ = 0.020 05 mol (3 significant figures + 1 guard digit)

If there is no volume change when the HCl dissolves in the water, the volume of the solution is 100 mL.

Molarity = $\frac{moles}{litres} = \frac{0.02005 mol}{0.100 L}$ $"moles"/"litres" = "0.02005 mol"/"0.100 L"$ = 0.200 mol/L

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If 500 mL of HCl gas at 300 K and 100 kPa dissolve in 100 mL of pure water, what is the concentration?

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