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Chemistry Question on Thermodynamics

One liter of water ( molecular weight 18.0618.06 ) weighs 0.9970 kg0.9970 \,kg . The degree of ionisation of water is............if Kw=1.10×10−14K_w = 1.10 \times 10^{-14} at 25∘C25^{\circ}C

A

1.05×10−71.05 \times 10^{-7}

B

1.9×10−91.9 \times 10^{-9}

C

1.01×10−111.01 \times 10^{-11}

D

4.52×10−74.52 \times 10^{-7}

Answer

1.9×10−91.9 \times 10^{-9}

Explanation

Solution

Given Kw=1.10×10−14K_w = 1.10 \times 10^{- 14} The value of KwK_w is temperature dependent as it is an equilibrium constant. The molarity of pure H2OH_2O =DensityM = \frac{\text{Density}}{M} =(1000 gL−118.0 g mol−1) = \left(\frac{1000\,gL^{-1}}{18.0\,g\,mol^{-1}}\right) =55.55 M = 55.55\,M α=10−7C\alpha = \frac{10^{-7}}{C} =10−755.6 = \frac{10^{-7}}{55.6} =1.9×10−9 = 1.9 \times 10^{-9}