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Q.

If N2 gas is bubbled through 1 L of water at 303K, how many moles of N2 gas exerts a partial pressure of 0.987 bar[Assuming henry’s law constant for N2 at 303K is 90K. bar]

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a

10.9×10−5

b

1.09×10−5

c

60.4×10−5

d

6.04×10−6

answer is B.

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Detailed Solution

PN2=KH×XN2

0.987 bar=90×103×xN2

xN2=nN2nN2+nH2O    nN2<<<nH2O

1.09×10−5=nN2nH2O     nH2O=100018

nN2=1.09×10−5×55.5

nN2=60.495×10−5

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