NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $$-$$4$$^\circ$$C before freezing. The amount of ice (in g) that will be separated out is __________. (Nearest integer)

[Given : Kf(H2O) = 1.86 K kg mol$$-$$1]

## Explanation

Let mass of water initially present = x gm

$$\Rightarrow$$ Mass of sucrose = (1000 $$-$$ x) gm

$$\Rightarrow$$ moles of sucrose = $$\left( {{{1000 - x} \over {342}}} \right)$$

$$\Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$$

$$\Rightarrow$$ 256.5x = 106 $$-$$ 1000x

$$\Rightarrow$$ x = 795.86 gm

$$\Rightarrow$$ moles of sucrose = 0.5969

New mass of H2O = a kg

$$\Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow$$ a = 0.2775 kg

$$\Rightarrow$$ ice separated = (795.86 $$-$$ 277.5) = 518.3 gm
2

### JEE Main 2021 (Online) 26th August Evening Shift

Numerical
83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ____________ K. (Nearest integer)

[Use : Molal Freezing point depression constant of water = 1.86 K kg mol$$-$$1]

Freezing Point of water = 273 K

Atomic masses : C : 12.0 u, O : 16.0 u, H : 1.0 u]

## Explanation

kf = 1.86 kg/mol

T$$_f^o$$ = 273 K

solvent : H2O(625 g)

Solute : 83 g of ethylene glycol

$$\Rightarrow$$ $$\Delta$$Tf = kf $$\times$$ m

$$\Rightarrow$$ $$\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$$

$$\Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}$$

$$\Rightarrow 273 - T_f^1 = 4$$

$$\Rightarrow T_f^1 = 259$$ K
3

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M C2C5OH is __________ (Integer answer)

(i) 0.10 M Ba3(PO4)2

(ii) 0.10 M Na2SO4

(iii) 0.10 M KCl

(iv) 0.10 M Li3PO4

## Explanation

As 0.1 M C2H5OH is non-dissociative and rest all salt given are electrolyte so in each case effective molarity > 0.1 so each will have lower freezing point.
4

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is __________ $$\times$$ 10$$-$$2.

(Round off to the nearest integer)

## Explanation Now, $$i = {{final\,moles} \over {initial\,moles}} = {{0.25a + 0.5a + 0.5a} \over {0.5a + 0.5a}}$$

= 1.25 = 125 $$\times$$ 10$$-$$2

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12