Question

Ratio of ∆Tb/kbof 10 gram of AB2and 14 g A2B in 100 gm of solvent in their respective solutions (ab2 and a2 b are non electrolyte) is 1 mole\ kg in both case hence atomic wt of a and b

Answer 1

Given : Ratio of $\frac{\Delta T_b}{K_b}$ of 10gm of AB₂ and 14gm of A₂B in 100gm of solvent in their respective solutions is 1 mole/kg in both cases.

To find : atomic weights of A and B.

solution : we know , $\Delta T_b=K_bm$

where m is molality of solution.

i.e., $\frac{\Delta T_b}{K_b}=m$

molality of AB₂ = {10/(A + 2B)}/100 × 1000

= 100/(A + 2B)

molality of A₂B = {14/(2A + B)}/100 × 1000

= 140/(2A + B)

Let $\frac{\Delta T_b}{K_b}$ = K

then, K₁ = m₁ and K₂ = m₂

⇒1 = {100/(A + 2B)}

⇒A + 2B = 100 ........(1)

K₂ = m₂

⇒1 = {140/(2A + B)}

⇒140 = 2A + B

⇒2A + B = 140 ..........(2)

from equations (1) and (2)

B = 20 and A = 60

Therefore atomic weights of A and B are 60 and 20 respectively.