Question

If $N_2$ gas is bubbled through water at ${20}^{0}C$, the volume of the gas dissolved in water at this temperature is:
(Given that total pressure at ${20}^{o}C=5.75\times {10}^7$ torr and partial pressure of $N_2$ at ${20}^{o}C=$ 742.5 torr)

• A. 32 ml
• B. 50 ml
• C. 16 ml
• D. 100 ml

Answer 1

The correct option is C 16 ml

The mole fraction of nitrogen $X_{N_2}=\frac{\text{partial pressure of}{N}_2}{\text{total pressure}}=\frac{742.5torr}{5.75\times {10}^{7}torr}=1.29\times {10}^{-5}$

55.5 moles of water are present in a liter of water. Let n be the number of moles of nitrogen. When n is small, the mole fraction of nitrogen will be $X_{N_2}=\frac{n}{n+55.5}=\frac{n}{55.5}$

Thus, $\frac{n}{55.5}=1.29\times {10}^{-5}$

The number of moles $n=55.5\times 1.29\times {10}^{-5}=7.16\times {10}^{-4}mol$

At STP, 1 litre of a gas occupies 22.4 L or 22400 ml.

Thus the volume of nitrogen at STP $=7.16\times {10}^{-4}\times 22400=16.0ml$.