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Chemistry Question on Solutions

The Henry�s law constant for O2O_2 dissolved in water is 4.34×1044.34 \times 10^4 atm at certain temperature. If the partial pressure of O2O_2 in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of O2O_2 in the solution?

A

1×10−51 \times 10^{-5}

B

1×10−41 \times 10^{-4}

C

2×10−52 \times 10^{-5}

D

1×10−61 \times 10^{-6}

Answer

1×10−51 \times 10^{-5}

Explanation

Solution

Given, KHK_{H} for O2O _{2} dissolved in water =4.34×104 atm=4.34 \times 10^{4} \,atm PO2=0.434 atmPO _{2} =0.434 \,atm x(x\left(\right. mole fraction of O2O _{2} in solution )=?)=? According to Henry's law, P=KH×xP =K_{H} \times x 0.4344.34×104=xO2\frac{0.434}{4.34 \times 10^{4}} =x_{ O _{2}} 1×10−5=xO21 \times 10^{-5} =x_{ O _{2}} ∴\therefore Mole fraction of O2O _{2} in water =1×10−5=1 \times 10^{-5}.