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Chemistry Question on Solutions

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

A

0.555

B

5.55

C

0.0555

D

55.5

Answer

0.555

Explanation

Solution

Answer (a) 0.555
According to the Question:-
Pgas=(KH)XgasP_{gas} = (K_{H})X_{gas}

1=(100×103)Xgas1= (100 \times 10^3)X_{gas}

Xgas=1×10−5X_{gas}=1\times 10^{-5}

ngasngas+nwater=1×10−5\frac{n_{gas}}{n_{gas} +n_{water}} = 1 \times 10^{-5}

ngas+nwater≈nwatern_{gas} +n_{water} ≈n_{water} (∵ngas<<<nwater)(∵ n_{gas} <<< n_{water})

ngas1000/18=1×10−5\frac{n_{gas}}{1000/18} = 1 \times 10^{-5}

ngas=5.556×10−4n_{gas} = 5.556 \times 10^{-4}

Therefore, the number of millimoles of the gas dissolved in one litre of water is ngas=0.555n_{gas} = 0.555