JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 21)
1 molal aqueous solution of an electrolyte A2B3 is 60% ionised. The boiling point of the solution at 1 atm is _________ K. (Rounded off to the nearest integer)
[Given Kb for (H2O) = 0.52 K kg mol$$-$$1]
[Given Kb for (H2O) = 0.52 K kg mol$$-$$1]
Answer
375
Explanation
$${A_2}{B_3}\buildrel {} \over
\longrightarrow 2{A^{ + 3}} + 3{B^{ - 2}}$$
No. of ions = 2 + 3 = 5
i = 1 + (n $$-$$ 1) $$\alpha $$
= 1 + (5 $$-$$ 1) $$\times$$ 0.6
= 1 + 4 $$\times$$ 0.6 = 1 + 2.4 = 3.4
$$\Delta {T_b} = {K_b} \times m \times i = 0.52 \times 1 \times 3.4 = 1.768^\circ $$ C
$$\Delta {T_b} = {({T_b})_{solution}} - {[{({T_b})_{{H_2}O}}]_{Solution}}$$
$$1.768 = {({T_b})_{solution}} - 100$$
$${({T_b})_{solution}} = 101.768^\circ $$ C
$$ = 375$$ K
No. of ions = 2 + 3 = 5
i = 1 + (n $$-$$ 1) $$\alpha $$
= 1 + (5 $$-$$ 1) $$\times$$ 0.6
= 1 + 4 $$\times$$ 0.6 = 1 + 2.4 = 3.4
$$\Delta {T_b} = {K_b} \times m \times i = 0.52 \times 1 \times 3.4 = 1.768^\circ $$ C
$$\Delta {T_b} = {({T_b})_{solution}} - {[{({T_b})_{{H_2}O}}]_{Solution}}$$
$$1.768 = {({T_b})_{solution}} - 100$$
$${({T_b})_{solution}} = 101.768^\circ $$ C
$$ = 375$$ K
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