Question

At ${20}^{o}C$ and a pressure of 1 atm each gas, 1 litre of water dissolves 0.043 g of pure $O_2$ or 0.019 g of pure $N_2$. Assuming that dry air is composed of 20% $O_2$ and 80% $N_2$ (by volume) determine the masses of $O_2$ and $N_2$ dissolved by 1 litre of water at ${20}^{o}C$ exposed to air at a total pressure of 1 atm. Assume V.P. of $H_2$ at ${20}^{o}C$ be 'b' atm

• A. $O_2=4.3\times {10}^{-3}g/L$,$NH2=0.75\times {10}^{-2}g/L$
• B. $O_2=5.25\times {10}^{-4}g/L$,$NH2=1.7\times {10}^{-3}g/L$
• C. $O_2=8.6\times {10}^{-3}g/L$,$NH2=1.5\times {10}^{-2}g/L$
• D. $O_2=10.5\times {10}^{-4}g/L$,$NH2=3.4\times {10}^{-3}g/L$

Answer 1

Answer: (C)

$O_2=8.6\times {10}^{-3}g/L$,$NH2=1.5\times {10}^{-2}g/L$

From the data given,

For pure $O_2:K_{H_{2}O}=\frac{a_{O_2}}{1}=0.04(P=1atm)$

For pure $N_2:K_{H_{N_2}}=\frac{a_{N_2}}{1}=0.019$

Now for air at 1 atm $P_{air}=P_{N_2}^{\prime }+P_{O_2}^{\prime }$

or $P_{O_2}^{\prime }=\frac{20}{100}\times 1$

$P_{N_2}^{\prime }=\frac{80}{100}\times 1$

For Pure $O_2:a_{O_2}=\frac{0.043\times 20}{100}=8.6\times {10}^{-3}g/L$

Fof Pure $NH2:a_{N_2}=\frac{0.019\times 80}{100}=1.5\times {10}^{-2}g/L$