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 % .

When the temperature of a solution increases then-

Molarity decreases and molality increases

Molality decreases and molarity increases

No change occur for both molarity and molality

Molarity increases and no change in molality.

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 12

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12th Assam Board

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Define the following term :

Ideal solution.

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