Solutions Question 248

Question: Henry’s law constant for N2 at 310 K is 82.35 kbar. N2 exerts a partial pressure of 0.840 bar. If N2 gas is bubbled through water at 293 K, then the number of millimoles of N2 that will dissolve in 1 L of water is

Options:

A) 0.0716

B) 1.30×105

C) 1.25×102

D) 0.0555

Show Answer

Answer:

Correct Answer: D

Solution:

Henry’s law constant is in the unit of pressure, hence we use relation p=KHχN2

χN2=pKH=0.840,bar82.35×103bar=1.0×105 1 L H2O=100mL,H2O

=1000,g,H2O(d=1,g,mL3)

Moles of H2O=100018=55.5 moles Let the number of moles of nitrogen be n Then χN2=nn+55.5n55.5 (since, n is very small)

n=55.5×1.0×105

=5.55×104mol=5.55×104×1000mmol

=5.55×102 millimole =0.0555 millimole



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