Two elements A and B form compounds of formula A B_2 and A B_4 . When dissolved in 20.0 ~g of benzene 1.0 ~g of A B_2 lowers f. pt. by 2.3^ C whereas 1.0 ~g of A B_4 lowers f . pt. by 1.3^ C . The K_f for benzene is

Correct answer: A. 25,42 Let the masses of A and B be a and b . The mass of A B_2 will be (a+2 b) g mol^ -1 and A B_4 will be (a+4 b) g mol^ -1 . Δ T_f=(K_f × W_B × 1000)/(M_B × W_A) On solving (i) and (ii), we get a=25.59 and b=42.64

AIIMS 2016

Exam: AIIMS
Year: 2016
Subject: Chemistry
Chapter: Solutions

AIIMS 2016 · Chemistry · Solutions

Two elements A and B form compounds of formula A B_2 and A B_4 ....

Two elements

A

and

B

form compounds of formula

A B_2

and

A B_4

. When dissolved in

20.0 \mathrm{~g}

of benzene

1.0 \mathrm{~g}

A B_2

lowers f. pt. by

2.3^{\circ} \mathrm{C}

whereas

1.0 \mathrm{~g}

A B_4

lowers

\mathrm{f}

. pt. by

1.3^{\circ} \mathrm{C}

. The

K_f

for benzene is 5.1. The atomic masses of

A

and

B

are

Two elements A and B form compounds of formula A B_2 and A B_4 . When dissolved in 20.0 ~g of benzene 1.0 ~g of A B_2

Options

A25,42
B42,25
C52,48
D48,52

Correct answer

A. 25,42

Step-by-step solution

Let the masses of

A

and

B

a

and

b

. The mass of

A B_2

will be

(a+2 b) \mathrm{g} \mathrm{mol}^{-1}

and

A B_4

will be

(a+4 b) \mathrm{g} \mathrm{mol}^{-1}

\Delta T_f=\frac{K_f \times W_B \times 1000}{M_B \times W_A}

On solving (i) and (ii), we get

a=25.59

and

b=42.64

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