JEE MAIN - Chemistry (2021 - 31st August Evening Shift - No. 15)
1.22 g of an organic acid is separately dissolved in 100 g of benzene (Kb = 2.6 K kg mol$$-$$1) and 100 g of acetone (Kb = 1.7 K kg mol$$-$$1). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by 0.17$$^\circ$$C. The increase in boiling point of solution in benzene in $$^\circ$$C is x $$\times$$ 10$$-$$2. The value of x is ______________. (Nearest integer) [Atomic mass : C = 12.0, H = 1.0, O =16.0]
Answer
13
Explanation
With benzene as solvent
$$\Delta$$Tb = i Kb m
$$\Delta$$Tb = $${1 \over 2}$$ $$\times$$ 2.6 $$\times$$ $${{1.22/{M_w}} \over {100/1000}}$$ .... (1)
With Acetone as solvent
$$\Delta$$Tb = i Kb m
0.17 = 1 $$\times$$ 1.7 $$\times$$ $${{1.22/{M_w}} \over {100/1000}}$$ ..... (2)
(1) / (2)
$${{\Delta {T_b}} \over {0.17}} = {{{1 \over 2} \times 2.6 + {{1.22/{M_w}} \over {100/1000}}} \over {1 \times 1.7 \times {{1.22/{M_w}} \over {100/1000}}}}$$
$$\Delta$$Tb = $${{0.26} \over 2}$$
$$\Delta$$Tb = 13 $$\times$$ 10$$-$$2
$$\Rightarrow$$ x = 13
$$\Delta$$Tb = i Kb m
$$\Delta$$Tb = $${1 \over 2}$$ $$\times$$ 2.6 $$\times$$ $${{1.22/{M_w}} \over {100/1000}}$$ .... (1)
With Acetone as solvent
$$\Delta$$Tb = i Kb m
0.17 = 1 $$\times$$ 1.7 $$\times$$ $${{1.22/{M_w}} \over {100/1000}}$$ ..... (2)
(1) / (2)
$${{\Delta {T_b}} \over {0.17}} = {{{1 \over 2} \times 2.6 + {{1.22/{M_w}} \over {100/1000}}} \over {1 \times 1.7 \times {{1.22/{M_w}} \over {100/1000}}}}$$
$$\Delta$$Tb = $${{0.26} \over 2}$$
$$\Delta$$Tb = 13 $$\times$$ 10$$-$$2
$$\Rightarrow$$ x = 13
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