JEE MAIN - Chemistry (2021 - 27th August Morning Shift - No. 20)
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)
[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
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
$$\Rightarrow$$ 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
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