The following questions are case-based questions. Each question has an internal choice and carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow.
The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid. According to Henry's law "the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution". Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry's law can be modified as "the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution:

(i) Henry's law constant for the solubility of methane in benzene at 298 K is 4.27 X 10⁵ mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is
(ii) The partial pressure of ethane over a saturated solution containing 6.56 x 10^-2 g of ethane is 1 bar. If the solution contains 5.00 x 10^-2 g of ethane then what will be the partial pressure (in bar) of the gas?
(iii) Which of the following statements is correct:
(a) K_H increases with increase of temperature
(b) K_H decreases with increase of temperature
(c) K_H remains constant with increase of temperature
(d) K_H first increases then decreases, with increase of temperature.
OR
18 g of glucose, $C_{6} H_{12} O_{6}$ (Molar mass – 180 g mol^-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (K_b for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K)

Question

The following questions are case-based questions. Each question has an internal choice and carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow.
The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid. According to Henry's law "the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution". Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry's law can be modified as "the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution:

(i) Henry's law constant for the solubility of methane in benzene at 298 K is 4.27 X 10⁵ mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is
(ii) The partial pressure of ethane over a saturated solution containing 6.56 x 10^-2 g of ethane is 1 bar. If the solution contains 5.00 x 10^-2 g of ethane then what will be the partial pressure (in bar) of the gas?
(iii) Which of the following statements is correct:
(a) K_H increases with increase of temperature
(b) K_H decreases with increase of temperature
(c) K_H remains constant with increase of temperature
(d) K_H first increases then decreases, with increase of temperature.
OR
18 g of glucose, $C_{6} H_{12} O_{6}$ (Molar mass – 180 g mol^-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (K_b for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K)

Infinity Learn · Accepted Answer

(i) KH​=4.27×105 mm Hg (at 298 K)p=760mmApplying Henry's law,p=KH​x, where x= Mole fraction or solubility of methane.x=​p​/KH =(760mm)​/(4.27×105mm)=1.78×10−3(ii) According to Henry's law, m = KH x P6.56×10−2=KHX1KH=6.56×10−2For another case,5×10−2=6.56×10−2×PP=56.56=0.762(iii) KH increases with increase of temperatureORWe know that : Elevation of boiling point ΔTb WBMB ×100× Kbwt. of solvent.Given: WB = 18 g MB = Formula of glucose is C6H12O6 = 6 × 12 + 12 + 6 × 16 = 180 Wt. of solvent = 1 kg or 1000 g, Kb = 0.52 K kg mol-1 Hence, ΔTb = 18g180×1000×0.521000g = 0.52 K ∴B.P of the solution = 373.15 + 0.052 = 373.202 K

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