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An ideal solution was prepared by dissol...

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.

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To solve the problem step by step, we will follow the outlined process in the video transcript, ensuring clarity and precision in each step. ### Step 1: Determine the Freezing Point Depression Constant (Kf) We start by calculating the freezing point depression constant (Kf) using the formula: \[ K_f = \frac{R \cdot T^2}{\Delta H_f \cdot 1000} \] Where: ...
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A solution is prepared by dissolving 10g of non-volatile solute in 200g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute. (Vapour pressure of pure water at 308K =32 mm Hg)

A solution is prepared by dissolving 9.25 g of non volatile solute in 450 ml of water. It has an osmotic pressure of 350 mm of Hg at 27^(@)C . Assuming the solute is non-electrolyte, determine its molecular mass. (R = 0.0821 lit atm K^(-1)mol^(-1) )

1000 g of 1 molal aqueous solution of sucrose is cooled and maintained at -3.534^(@)C . Find out how much ice will separate out at this temperature. (K_(f) for water =1.86 k m^(-1) )

A one litre solution is prepared by dissolving some lead-nitrate in water. The solution was found to boil at 100.15^(@)C . To the resulting solution 0.2 mole NaCI was added. The resulting solution was found to freeze at 0.83^(@)C . Determine solubility product of PbCI_(2) . Given K_(b) = 0.5 and K_(f) = 1.86 . Assume molality to be equal to molarity in all case.

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)

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A thermally isolated vessel contains 100 g of water at 0^(@)C when air above the water is pumped out, some of the water freezes and some evaporates at 0^(@)C itself. Calculate the mass at 0^(@)C=2.10xx10^(6) j//kg and latent heat of fusion of ice =3.36xx10^(5) j//kg .

(a) A solution is prepared by dissolving 10g of non-volatile solute in 180g of H_(2)O . If the relative lowering of vapour pressure is 0.005 , find the molar mass of the solute. (b) How many grams of sucrose (C_(12)H_(22)O_(11)) must be dissolved in 90g of water to produce a solution over which the relative humidity is 80% ? Assume the solution is ideal.

A solution containing 0.1 mol of naphthalene and 0.9 mol of benzene is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed upto 353 K where its vapour pressure was found to be 670 mm . The freezing point and boiling point of benzene are 278.5 K and 353 K respectively, and its enthalpy of fusion is 10.67 KJ "mol"^(-1) . Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume ideal behaviour.

Properties such as boiling point, freezing point and vapour, pressure of a pure solvent change Propeties such as boiling point, freezing point and vapour, pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 Given Freezing point depression constant of water (K_(f)^("water"))=1.86 K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol"))=2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water"))=0.52 kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol"))=1.2 kg "mol"^(-1) Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water =373 K tandard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mmHg Vapour presure of pure ethanol =40g Hg Molecular weight of water =18 g"mol"^(-1) Molecules weight of ethanol =46 g "mol"^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 The freezing point of the solution M is

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