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 $^{\circ}\mathrm{C}$ of the system on mixing is $x \times 10^{-2}$. The value of $x$ is ______. (Nearest integer) [Given : Specific heat of water $=4.18 \mathrm{~J~g}^{-1} \mathrm{~K}^{-1}$ Density of water $=1.00 \mathrm{~g~cm}^{-3}$] (Assume no volume change on mixing)

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Chemical Thermodynamics

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 1 0^{- 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)

Answer: 82

The acid supplies

0.2 \times 0.2 = 0.04

mol

H^{+}

and the base

0.1 \times 0.3 = 0.03

mol

O H^{-}

, so the base is limiting and

0.03

mol of water forms, releasing

0.03 \times 57.1 = 1.713 kJ = 1713 J

. The mixture occupies

500 mL

with mass

500 g

, so

Δ T = \frac{1713}{500 \times 4.18} = 0.82^{\circ} C

. With

Δ T = x \times 1 0^{- 2}

, the value of

x

82

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Integer answer