200 , ~mL of 0.2 , M HCl is mixed with 300 , ~mL of 0.1 , M NaOH.… - Vidyadip

MCQ

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

[Given : Specific heat of water $=4.18\, \mathrm{~J} \,\mathrm{~g}^{-1}\, \mathrm{~K}^{-1}$

Density of water $=1.00\, \mathrm{~g}\, \mathrm{~cm}^{-3}$ ]

(Assume no volume change on mixing)

A
$12$
B
$125$
✓
$82$
D
$74$

✓

Answer

Correct option: C.

$82$

c
$\Rightarrow \text { Millimoles of } \mathrm{HCl}=200 \times 0.2=40$

$\Rightarrow \text { Millimoles of } \mathrm{NaOH}=300 \times 0.1=30$

$\Rightarrow \text { Heat released }=\left(\frac{30}{1000} \times 57.1 \times 1000\right)=1713 \,\mathrm{~J}$

$\Rightarrow \text { Mass of solution }=500 \,\mathrm{ml} \times 1 \,\mathrm{gm} / \mathrm{ml}=500 \,\mathrm{gm}$

$\Rightarrow \Delta \mathrm{T}=\frac{\mathrm{q}}{\mathrm{m} \times \mathrm{C}}=\frac{1713\, \mathrm{~J}}{500\, \mathrm{~g} \times 4.18\, \frac{\mathrm{J}}{\mathrm{g}-\mathrm{K}}}=0.8196\, \mathrm{~K}$

$=81.96 \times 10^{-2}\, \mathrm{~K}$

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