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JEE Main 2022 · 24 Jun · Shift 2, Chemistry question 82

A company dissolves 'x' amount of CO2\mathrm{CO_2} at 298 K in 1 litre of water to prepare soda water. X = ______ ×10−3\times10^{-3} g. (nearest integer)

(Given: partial pressure of CO2\mathrm{CO_2} at 298 K = 0.835 bar.

Henry's law constant for CO2\mathrm{CO_2} at 298 K = 1.67 kbar.

Atomic mass of H, C and O is 1, 12, and 6 g mol−1^{-1}, respectively)

Answer and worked solution

Official answer: 1221

Henry's law. xCO2=pKH=0.8351670=5×10−4x_{\mathrm{CO_2}}=\dfrac p{K_H}=\dfrac{0.835}{1670}=5\times10^{-4}.

Moles. 1 L of water is 100018=55.6\frac{1000}{18}=55.6 mol, so nCO2≈5×10−4×55.6≈0.0278n_{\mathrm{CO_2}}\approx5\times10^{-4}\times55.6\approx0.0278 mol.

Mass. 0.0278×44≈1.2220.0278\times44\approx1.222 g =1222×10−3=1222\times10^{-3} g. The paper prints the atomic mass of O as 6, a misprint for 16; the key accepts 1221 or 1222.

Answer: 1222

Topics:Concentration and solubility
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