The freezing point of 3.75xx10^(-2)M aqu...

The freezing point of `3.75xx10^(-2)M` aqueous solution of weak acid HA is `272.9K`. The molality of the solution was found to be `0.0384` molal. Find the `[H^(+)]` of the solution on adding `3.75xx10^(-2)` moles of NaA to one litre of the above solution. `(K_(f) "of water" =1.86 "molal"^(-1))`

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The correct Answer is:

`6.89xx10^(-3)M;`

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The freezing point of 0.20 M solution of week acid HA is 272.5 K . The molality of the solution is 0.263 "mol" kg^(-1) . Find the pH of the solution on adding 0.25 M sodium acetate solution. K_(f) of water = 1.86 K m^(-1)

The freezing point of 0.20M solution of weak acid HA is 272.5K . The molality of the solution is 0.263mol Kg^(-1) . a. Find the pH of the solution on adding 0.25m solution of acetate of the above solution. b. Find the pH of the solution on adding 0.20M solution of NaOH . Given: K_(f) of water = 1.86 Km^(-1)

The freezing point of an aqueous solutions is 272 .93 K . Calculate the molality of the solution if molal depression constant for water is 1.86 Kg mol^(-1)

A 0.2 molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of the solution is (k_(f) = 1.86 K kg "mole"^(-1) for water):

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If 0.2 molal aqueous solution of a weak acid (HA) is 40% ionised then the freezing point of the solution will be ( K_f for water = 1.86^@ C

The freezing point of a 0.05 molal solution of a non-electrolyte in water is: ( K_(f) = 1.86 "molality"^(-1) )

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