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Choose the Best Answer · Q5

Q.The Henry's law constant for the solubility of Nitrogen gas in water at 350 K is 8×1048 \times 10^{4} atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is

(a) 4×10−44 \times 10^{-4}
(b) 4×1044 \times 10^{4}
(c) 2×10−22 \times 10^{-2}
(d) 2.5×10−42.5 \times 10^{-4}
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Step 1. Total pressure is 4 atm and N2_2's mole fraction in air is 0.5, so the partial pressure of N2_2 is pN2=0.5×4=2p_{N_2}=0.5\times4=2 atm.

Step 2. By Henry's law, the mole fraction of N2_2 dissolved in water is xN2=pN2KH=28×104=2.5×10−5x_{N_2}=\dfrac{p_{N_2}}{K_H}=\dfrac{2}{8\times10^{4}}=2.5\times10^{-5}. …

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