One liter of water ( molecular weight 18.06 ) weighs 0.9970 kg . The degree of ionisation of water is............if Kw = 1.10 × 10-14 at 25°C

Question

One liter of water ( molecular weight 18.06 ) weighs 0.9970 kg . The degree of ionisation of water is............if Kw = 1.10 × 10-14 at 25°C  (A)  1.05 × 10-7  (B)  1.9 × 10-9  (C)  1.01 × 10-11  (D)  4.52 × 10-7

Tardigrade Admin · Accepted Answer

Given Kw = 1.10 × 10- 14 The value of Kw is temperature dependent as it is an equilibrium constant. The molarity of pure H2O = ( textDensity/M) = ((1000 gL-1/18.0 g mol-1)) = 55.55 M α = (10-7/C) = (10-7/55.6) = 1.9 × 10-9

Q. One liter of water ( molecular weight $18.06$ ) weighs $0.9970 k g$ . The degree of ionisation of water is............if $K_{w} = 1.10 \times 1 0^{- 14}$ at $2 5^{\circ} C$

$1.05 \times 1 0^{- 7}$

$1.9 \times 1 0^{- 9}$

$1.01 \times 1 0^{- 11}$

$4.52 \times 1 0^{- 7}$

Q. One liter of water ( molecular weight 18.06 ) weighs 0.9970kg . The degree of ionisation of water is............if Kw​=1.10×10−14 at 25∘C

1.05×10−7

1.9×10−9

1.01×10−11

4.52×10−7

Given Kw​=1.10×10−14 The value of Kw​ is temperature dependent as it is an equilibrium constant. The molarity of pure H2​O =MDensity​ =(18.0gmol−11000gL−1​) =55.55M α=C10−7​ =55.610−7​ =1.9×10−9

Q. One liter of water ( molecular weight $18.06$ ) weighs $0.9970 k g$ . The degree of ionisation of water is............if $K_{w} = 1.10 \times 1 0^{- 14}$ at $2 5^{\circ} C$

$1.05 \times 1 0^{- 7}$

$1.9 \times 1 0^{- 9}$

$1.01 \times 1 0^{- 11}$

$4.52 \times 1 0^{- 7}$