JEE Challenger
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ChemMedAdv 2009 • Paper 1

Solubility of Nitrogen Gas in Water Using Henrys Law

The Henry's law constant for the solubility of N2\text{N}_2 gas in water at 298 K298\text{ K} is 1.0×105 atm1.0 \times 10^5\text{ atm}. The mole fraction of N2\text{N}_2 in air is 0.80.8. The number of moles of N2\text{N}_2 from air dissolved in 10 moles10\text{ moles} of water at 298 K298\text{ K} and 5 atm5\text{ atm} pressure is

Options

A

4.0×10−44.0 \times 10^{-4}

Correct
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}

Topics & Concepts

SolutionsConcentration Terms

Step-by-Step Solution

To find the number of moles of N2\text{N}_2 gas dissolved in water, we apply Henry's law.

  1. Calculate the partial pressure of N2\text{N}_2 in air:
    According to Dalton's law of partial pressures:
    pN2=xN2(air)×Ptotalp_{\text{N}_2} = x_{\text{N}_2(\text{air})} \times P_{\text{total}}
    Given:

    • Mole fraction of N2\text{N}_2 in air, xN2(air)=0.8x_{\text{N}_2(\text{air})} = 0.8
    • Total pressure, Ptotal=5 atmP_{\text{total}} = 5\text{ atm}

    pN2=0.8×5 atm=4.0 atmp_{\text{N}_2} = 0.8 \times 5\text{ atm} = 4.0\text{ atm}

  2. Calculate the mole fraction of N2\text{N}_2 dissolved in water:
    According to Henry's law:
    pN2=KH×xN2(aq)p_{\text{N}_2} = K_{\text{H}} \times x_{\text{N}_2(\text{aq})}
    Given:

    • Henry's law constant, KH=1.0×105 atmK_{\text{H}} = 1.0 \times 10^5\text{ atm}

    xN2(aq)=pN2KH=4.0 atm1.0×105 atm=4.0×10−5x_{\text{N}_2(\text{aq})} = \frac{p_{\text{N}_2}}{K_{\text{H}}} = \frac{4.0\text{ atm}}{1.0 \times 10^5\text{ atm}} = 4.0 \times 10^{-5}

  3. Calculate the number of moles of N2\text{N}_2 dissolved:
    The mole fraction in the solution is given by:
    xN2(aq)=nN2nN2+nH2Ox_{\text{N}_2(\text{aq})} = \frac{n_{\text{N}_2}}{n_{\text{N}_2} + n_{\text{H}_2\text{O}}}

    Since the solubility of nitrogen gas is very small, nN2≪nH2On_{\text{N}_2} \ll n_{\text{H}_2\text{O}}, so:
    xN2(aq)≈nN2nH2Ox_{\text{N}_2(\text{aq})} \approx \frac{n_{\text{N}_2}}{n_{\text{H}_2\text{O}}}

    Given that nH2O=10 molesn_{\text{H}_2\text{O}} = 10\text{ moles}:
    nN2=xN2(aq)×nH2O=4.0×10−5×10=4.0×10−4 molesn_{\text{N}_2} = x_{\text{N}_2(\text{aq})} \times n_{\text{H}_2\text{O}} = 4.0 \times 10^{-5} \times 10 = 4.0 \times 10^{-4}\text{ moles}

Thus, the number of moles of N2\text{N}_2 dissolved is 4.0×10−44.0 \times 10^{-4}.

Therefore, the correct option is A.

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