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NEET CHEMISTRYMedium

The Henry's law constant for the solubility of N2N_2 gas in water at 298 K298 \text{ K} is 1.0×105 atm1.0 \times 10^5 \text{ atm}. The mole fraction of N2N_2 in air is 0.80.8. The number of moles of N2N_2 formed from air dissolved in 10 moles10 \text{ moles} of water at 298 K298 \text{ K} and 5 atm5 \text{ atm} pressure is:

A

4.0 × 10⁻⁴ mol

B

4.0 × 10⁻⁵ mol

C

5.0 × 10⁻⁴ mol

D

4.0 × 10⁻⁶ mol

Step-by-Step Solution

  1. Calculate Partial Pressure of Nitrogen (pN2p_{N_2}): According to Dalton's Law, the partial pressure of a component in a gas mixture is the product of its mole fraction in the gas phase and the total pressure. pN2=Mole fraction in air×Ptotalp_{N_2} = \text{Mole fraction in air} \times P_{\text{total}} pN2=0.8×5 atm=4.0 atmp_{N_2} = 0.8 \times 5 \text{ atm} = 4.0 \text{ atm}

  2. Calculate Mole Fraction of Nitrogen in Solution (xN2x_{N_2}): Using Henry's Law formula: p=KH×xp = K_H \times x 4.0 atm=(1.0×105 atm)×xN24.0 \text{ atm} = (1.0 \times 10^5 \text{ atm}) \times x_{N_2} xN2=4.01.0×105=4.0×105x_{N_2} = \frac{4.0}{1.0 \times 10^5} = 4.0 \times 10^{-5}

  3. Calculate Moles of Nitrogen (nN2n_{N_2}): The mole fraction xN2x_{N_2} is defined as nN2nN2+nH2O\frac{n_{N_2}}{n_{N_2} + n_{H_2O}}. For dilute solutions, nN2nH2On_{N_2} \ll n_{H_2O}, so we approximate xN2nN2nH2Ox_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}}. 4.0×105=nN2104.0 \times 10^{-5} = \frac{n_{N_2}}{10} nN2=4.0×105×10=4.0×104 moln_{N_2} = 4.0 \times 10^{-5} \times 10 = 4.0 \times 10^{-4} \text{ mol}

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