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1

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

Numerical
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is $$-$$57.1 kJ. The increase in temperature in $$^\circ$$C of the system on mixing is x $$\times$$ 10$$-$$2. The value of x is ___________. (Nearest integer)

[Given : Specific heat of water = 4.18 J g$$-$$1 K$$-$$1, Density of water = 1.00 g cm$$-$$3]

[Assume no volume change on mixing)

## Explanation

$$\Rightarrow$$ Millimoles of HCl = 200 $$\times$$ 0.2 = 40

$$\Rightarrow$$ Millimoles of NaOH = 300 $$\times$$ 0.1 = 30

$$\Rightarrow$$ Heat released = $$\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right)$$ = 1713 J

$$\Rightarrow$$ Mass of solution = 500 ml $$\times$$ 1 gm/ml = 500 gm

$$\Rightarrow$$ $$\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}}$$ = 0.8196 K

= 81.96 $$\times$$ 10$$-$$2 K
2

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

Numerical
For water $$\Delta$$vap H = 41 kJ mol$$-$$1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___________ kJ mol$$-$$1

[Use : R = 8.3 J mol$$-$$1 K$$-$$1]

## Explanation

H2O(l) $$\to$$ H2O(g) : $$\Delta$$H = 41 $${{kJ} \over {mol}}$$

$$\Rightarrow$$ From the relation : $$\Delta$$H = $$\Delta$$U + $$\Delta$$ngRT

$$\Rightarrow$$ 41$${{kJ} \over {mol}}$$ = $$\Delta$$U + (1) $$\times$$ $${{8.3} \over {1000}}$$ $$\times$$ 373

$$\Delta$$ DU = 41 $$-$$ 3.0959 = 38 kJ/mol
3

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

Numerical
The Born-Haber cycle for KCl is evaluated with the following data :

$${\Delta _f}{H^\Theta }$$ for KCl = $$-$$436.7 kJ mol$$-$$1 ;

$${\Delta _{sub}}{H^\Theta }$$ for K = 89.2 kJ mol$$-$$1 ;

$${\Delta _{ionization}}{H^\Theta }$$ for K = 419.0 kJ mol$$-$$1 ;

$${\Delta _{electron\,gain}}{H^\Theta }$$ for Cl(g) = $$-$$348.6 kJ mol$$-$$1 ;

$${\Delta _{bond}}{H^\Theta }$$ for Cl2 = 243.0 kJ mol$$-$$1

The magnitude of lattice enthalpy of KCl in kJ mol$$-$$1 is _____________ (Nearest integer)

## Explanation

$${\Delta _f}H_{KCl}^\Theta = {\Delta _{sub}}H_{(K)}^\Theta + {\Delta _{ioniztion}}H_{(K)}^\Theta + {1 \over 2}{\Delta _{bond}}H_{(C{l_2})}^\Theta + {\Delta _{electron\,gain}}H_{(Cl)}^\Theta + {\Delta _{lattice}}H_{(KCl)}^\Theta$$

$$\Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta$$

$$\Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$$ kJ mol$$-$$1

The magnitude of lattice enthalpy of KCl in kJ mol$$-$$1 is 718 (Nearest integer).
4

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

Numerical
When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is __________ $$\times$$ 10$$-$$2 K. (Round off to the nearest integer).

[Use : H+ (aq) + OH$$-$$ (aq) $$\to$$ H2O : $$\Delta$$$$\gamma$$H = $$-$$57.1 kJ mol$$-$$1]

Specific heat of H2O = 4.18 J K$$-$$1 g$$-$$1

density of H2O = 1.0 g cm$$-$$3

Assume no change in volume of solution on mixing.

## Explanation

$${n_{{H^ + }}} = {{400 \times 0.2} \over {1000}} \times 2 = 0.16$$

$${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$$ (L.R.)

Now, heat liberated from reaction = heat gained by solutions

or, 0.06 $$\times$$ 57.1 $$\times$$ 103

= (1000 $$\times$$ 1.0) $$\times$$ 4.18 $$\times$$ $$\Delta$$T

$$\therefore$$ $$\Delta$$T = 0.8196K

= 81.96 $$\times$$ 10$$-$$2 K $$\approx$$ 82 $$\times$$ 10$$-$$2 K

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