1 . 00 g of a non-electrolyte solute dissolved in 50 . 0 g of benzene lowered the freezing point of benzene by 0 . 40 K . The freezing point depression constant of benzene is 5 . 12 K Kg mol - 1 . Find molar mass of solute.

As ∆ T f = K f × m where, K f = Freezing point depression constant ∆ T f = Depression in freezing point, m = molality Now, The molar mass is the molecular weight, It is the sum of the total mass in grams of all the atoms that make up a mole of a particular molecule. ∆ T f = K f × W 2 × 1000 M 2 × W 1 where, W 2 = mass of solute, W 1 = the mass of benzene solvent, M 2 = Molar mass of solute = 0 . 40 = 5 . 12 × 1 . 0 × 1000 M 2 × 50 = M 2 = 256 g mol - 1

Calculate the mass of a compound (molar mass = 256 g mol - 1 ) to be dissolved in 75 g of benzene to lower its freezing point by 0 . 48 K . ( K f = 5 . 12 K kg mol - 1 )

Given, The molar mass of the compound M B = 256 g mol - 1 The weight of the benzene W A = 75 g Depression in freezing point ∆ T f = 0 . 48 K Molal freezing point depression constant K f = 5 . 12 K Kg mol - 1 If the mass of the compound is W B then, ⇒ W B = ∆ T f × W A × M B K f × 1000 = 0 . 48 × 75 × 256 5 . 12 × 1000 = 1 . 8 g The mass of compound is 1 . 8 g .

What is meant by positive deviations from Raoult's law and how is the sign of ∆ H mix related to positive deviation from Raoult's law?

Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆ H mix is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy and less energy is released on formation of weaker interactions. So overall ∆ H mix is positive.

Define molality of a solution.

Molality (m) of a solution is defined as the number of moles of solute in 1 Kg of the solvent. The molality of a solution does not change with the change in temperature of the solution. Molality ( m ) = Moles of solute Mass of solvent in kg

1 . 00 molal aqueous solution of trichloroacetic acid CCl 3 COOH is heated to its boiling point. The solution has a boiling point of 100 . 18 ° C . Determine the van’t Hoff factor for trichloroacetic acid ( K b for water = 0 . 512 K kg mol - 1 ).

Van't Hoff factor is used to express the extent of dissociation or association of the solute in solution. CCl 3 COOH → CCl 3 COO - + H + In order to calculate Van’t Hoff factor, we need to first calculate the change in boiling point that is ΔT b . The change in boiling point is calculated by using the following expression. ∆ T b = The boiling point of the solution - Boiling point of the pure solvent ∆ T b = ( 100 . 18 + 273 ) - ( 100 + 273 ) = 0 . 18 K . Here , the molality is 1 . Hence , ∆ T b = iK b m = iK b × 1 i = 0 . 18 0 . 512 i = 0 . 35 .

A solution is obtained by mixing 300g of 25% solution and 400g of 40% solution by mass. Calculate the mass percentage (w/w) of water in solution.

Given, 300 g of 25 % solution contains 300 × 25 100 = 75 g of solute 400 g of 40 % solution contains 400 × 40 100 = 160 g of solute Therefore , the total mass of the solute is = 75 g of solute + 160 g of solute = 235 g of solute Total mass of the solution = 400 + 300 = 700 g Hence , the mass percentage w / w of water = Mass of solute Mass of the solution × 100 = 235 700 × 100 = 33 . 5 % .

Define the following term : Ideal solution.

Ideal solution.: It is defined as the solution in which the interaction between component molecules does not normally vary from the interactions between the molecules of each component or the solution that exactly conforms to Raoult's law is called an ideal solution.

Define the following term : Van't Hoff factor

Van't Hoff factor: It is defined as the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to calculated colligative property.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Solutions>Assam Board-2020>Q 8

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