1000 g H2O have 0.1 mole urea and it's freezing point is -0.2 degree now it's freezed up to -2 degree then how much amount of ice will form

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Q. Depression of freezing point of 0.01 molal aqueous Acetic Acid solution is 0.0204 degree Celsius. 1 molal urea solution freezes at -1.86 degree Celsius. Assuming molality equal to molarity, pH of Acetic Acid solution.

Q. A solution of x moles of sucrose in 100 grams of water freezes at

- {0.2}^{o} C

. As ice separates the freezing point goes down to

{0.25}^{o} C

. How many grams of ice would have separated?

0.1 m

urea solutions will have freezing point

(K_{f} = 1.86 K k g m o l^{- 1})

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1000 g H2O have 0.1 mole urea and it's freezing point is -0.2 degree now it's freezed up to -2 degree then how much amount of ice will form​

1000 g H2O have 0.1 mole urea and it's freezing point is -0.2 degree now it's freezed up to -2 degree then how much amount of ice will form