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Chemistry Question on Solutions

When a gas is bubbled through water at 298 K298 \,K, a very dilute solution of gas is obtained. Henry?? law constant for the gas is 100 kbar100\, kbar. If gas exerts a pressure of 1 bar1\, bar, the number of moles of gas dissolved in 11 litre of water is

A

0.5550.555

B

55.55×10−555.55 \times 10^{-5}

C

55.55×10−355.55 \times 10^{-3}

D

5.55×10−55.55 \times 10^{-5}

Answer

55.55×10−555.55 \times 10^{-5}

Explanation

Solution

p=KH×xp = K_{H} \times x x=pKH=1100×103x = \frac{p}{K_{H}} = \frac{1}{100 \times 10^{3}} =1×10−5= 1 \times 10^{-5} Mole fraction =Moles of gasTotal moles= \frac{\text{Moles of gas}}{\text{Total moles}} Moles of H2O=100018H_{2}O = \frac{1000}{18} =55.55(∵1 L=1000 g)= 55.55 \left(\because 1 \,L = 1000\,g\right) Mole fraction =xx+55.55 (55.55>>>x)= \frac{x}{x+55.55}\,\left(55.55 > > > x\right) ∴10−5=x55.55\therefore 10^{-5} = \frac{x}{55.55} or x=55.55×10−5x = 55.55 \times 10^{-5}