JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 18)

Molecules of benzoic acid (C₆H₅COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that K_f = 5 K kg mol^–1, Molar mass of benzoic acid = 122 g mol^–1)

1.5 g

1.8 g

1.0 g

2.4 g

Explanation

We know,

$$\Delta $$T_f = i K_f $$ \times $$ $${w \over M} \times {{1000} \over {{w_s}}}$$

i = 1 + $$\alpha $$$$\left( {{1 \over n} - 1} \right)$$

Here Benzoic acid dimerise, so value of n = 2

$$ \therefore $$  i = 1 + 0.8 $$\left( {{1 \over 2} - 1} \right)$$

= 1 $$-$$ 0.4

= 0.6

$$ \therefore $$  2 = 0.6 $$ \times $$ 5 $$ \times $$ $${w \over {122}} \times {{1000} \over {30}}$$

$$ \Rightarrow $$  w = 2.44 g

JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 18)

Explanation

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