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Choose the Best Answer · Q11

Q.The Henry's law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B (in the gas phase) is 0.2. The ratio of mole fraction of B to A dissolved in water will be

(a) xy×2\dfrac{x}{y} \times 2
(b) yx×0.2\dfrac{y}{x} \times 0.2
(c) xy×0.2\dfrac{x}{y} \times 0.2
(d) xy×5\dfrac{x}{y} \times 5
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Step 1. Let the gas-phase mole fractions of A and B be yAy_A and yBy_B, with yAyB=0.2\dfrac{y_A}{y_B}=0.2 (given). Their partial pressures above the solution are pA=yAPp_A=y_A P and pB=yBPp_B=y_B P (P = total pressure), so pApB=yAyB=0.2\dfrac{p_A}{p_B}=\dfrac{y_A}{y_B}=0.2.

Step 2. By Henry's law, the mole fraction each gas actually dissolves to is xA(dissolved)=pAKH,A=pAxx_A(\text{dissolved})=\dfrac{p_A}{K_{H,A}}=\dfrac{p_A}{x} and xB(dissolved)=pBKH,B=pByx_B(\text{dissolved})=\dfrac{p_B}{K_{H,B}}=\dfrac{p_B}{y} (using the problem's own symbols xx, yy for the two Henry's constants). …

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