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Chemistry Question on Colligative Properties

The Henry?? law constant for the solubility of N2N_2 gas in water at 298 K298\, K is 1.0×105 atm1.0 \times 10^5\, atm. The mole fraction of N2N_2 in air is 0.80.8. The number of moles of N2N_2 from air dissolved in 1010 moles of water at 298 K298\, K and 5 atm5 \,atm pressure is

A

4.0×10−44.0 \times 10^{-4}

B

4.0×10−54.0 \times 10^{-5}

C

5.0×10−45.0 \times 10^{-4}

D

4.0×10−64.0 \times 10^{-6}

Answer

4.0×10−44.0 \times 10^{-4}

Explanation

Solution

According to Henry?? law, xN2×KH=pN2x_{N_2} \times K_H = p_{N_2} (pN2=p_{N_2} = Partial pressure of N2N_2) Given, total pressure =5 atm= 5\, atm mole fraction of N2=0.8N_2 = 0.8 ∴ \therefore Partial pressure of N2=0.8×5=4N_{2} = 0.8 \times 5 = 4 ⇒xN2×1×105=4\Rightarrow x_{N_2} \times 1 \times 10^{5} = 4 ⇒xN2=4×10−5\Rightarrow x_{N_2} = 4 \times 10^{-5} no. of moles of H2OH_{2}O, nH2O=10n_{H_2O} = 10 no. of nmoles of N2N_{2}, nN2=?n_{N_2 }= ? nN2nN2+nH2O=xN2=4×10−5\frac{n_{N_2}}{n_{N_2}+ n_{H_2O}} = x_{N_2}=4 \times 10^{-5} ⇒nN210+nN2=4×10−5\Rightarrow \frac{n_{N_2}}{10 +n_{N_2}} = 4\times10^{-5} ⇒nN2=4×10−4[∵nN2<<<10]\Rightarrow n_{N_2} = 4 \times 10^{-4} \quad\left[\because n_{N_2} <<< 10\right]