IIT-JEE 2009 · Paper 1 · Chemistry Q4

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ChemistrySingle correct+3 / -1Question 4 of 60

The Henry's law constant for the solubility of $\ce{N2}$ gas in water at 298 K is $1.0\times10^5$ atm. The mole fraction of $\ce{N2}$ in air is 0.8. The number of moles of $\ce{N2}$ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

4.0\times10^{-4}

4.0\times10^{-5}

5.0\times10^{-4}

4.0\times10^{-6}

Answer and worked solution

Official answer: A

Partial pressure. $p_{\ce{N2}}=0.8\times5=4$ atm.

Henry's law. $x_{\ce{N2}}=\dfrac{p}{K_H}=\dfrac4{1.0\times10^5}=4\times10^{-5}$ .

Moles. For a dilute solution $x\approx\dfrac{n}{n_\text{water}}$ , so $n=4\times10^{-5}\times10=4.0\times10^{-4}$ mol.

(A)

Topics:Concentration and solubility

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IIT-JEE 2009 · Paper 1, Chemistry question 4