[" Tervagr for wolatiel in us moles of water."],[[" (271K) where some ice get separated out."],[" 700tor at 373K.Calculate the mass of ice "]]

An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in 0.9 moles of water. The solution was then cooled just below its freezing temperature (271K) , where some ice get separated out. The remaining aqueous solution registered a vapour pressure of 700 torr at 373 K . Calculate the mass of ice separated out, if the molar heat of fusion of water is 96 kJ .

An ideal solution was prepared by dissolving some amount of can sugar (non-volatile) in 0.9 moles of water.The solution was then cooled just below its freezing temperature (271 K) where some ice get separated out.The remaining aqueous solution registered a vapour pressure of 700 torr at 373K.Calculate the mass of ice separated out, if the molar heat of fusion of water is 6 kJ.

a) 24 g of a non-volatile, non-electrolyte solute is added to 600 g of water. The boiling point of the resulting solution is 373.35K. Calculate the molar mass of the solute ("Given boiling point of pure water = 373 K and Kb for water=0.52 K kg mol"^(-1)) .

Why is ice at 273 K more effective in cooling than water at the some temperature?

Assertion:When a solution of non-volatile solute in water is cooled slightly below its freezing temperature, some ice separates out and then freezing stops. Reason: Separation of ice increases the molality of the left over solution.

Calculate the change in entropy for the fusion of 1 mole of ice (water). The melting point of water is 273 K and molar enthalpy of funsion for water =6.0 kJ mol^(-1)

Calculate the change in entropy for the fusion of 1 mole of ice (water). The melting point of water is 273 K and molar enthalpy of funsion for water =6.0 kJ mol^(-1)

2 moles of non - volatile solute is added to 1 kg water at -8^(@)C. K_(f) of water is 2 "K kg mol"^(-1) . Mass of ice that separates out is (Ignoring the effect of change in volume)