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1

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is __________ K. (Nearest integer) [Given : Kf = 1.86 K kg mol$$-$$1; Density of water = 1.00 g cm$$-$$3; Freezing point of water = 273.15 K]
Your Input ________

Answer

Correct Answer is 271

Explanation

Molality = $${{\left( {{{40} \over {180}}} \right)mol} \over {0.2\,Kg}} = \left( {{{10} \over 9}} \right)$$ molal

$$ \Rightarrow \Delta {T_f} = {T_f} - {T_f}' = 1.86 \times {{10} \over 9}$$

$$ \Rightarrow {T_f}' = 273.15 - 1.86 \times {{10} \over 9}$$

= 271.08 K

$$ \simeq $$ 271 K (nearest - integer)
2

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]
Your Input ________

Answer

Correct Answer is 518

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
3

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]
Your Input ________

Answer

Correct Answer is 269

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
4

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
Your Input ________

Answer

Correct Answer is 4

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.

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