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```
ab2 and ab4 when dissolved in 20gms of benzene 1 gm of ab2 delta Tf by 2.4k where on 1gm ab4
ab2 and ab4 when dissolved in 20gms of benzene 1 gm of ab2 delta Tf by 2.4k where on 1gm ab4

```
2 years ago

## Answers : (1)

Arun
25768 Points
```							As We know that: MB = (Kf x wB x 1000) / (wA X ΔTf) Now ΔTf = 2.3 , wB = 1.0 , wA = 20, KF = 5.1 (given) PUTTING THE VALUES IN THE EQUATION MB = (5.1 x 1 x 1000) / (20 x 2.3) = 110.87 g/mol Therefore MAB2 = 110.9 For AB4 compound ΔTf = 1.3 , wb = 1 ,wa = 20 MB = (5.1 X 1 X 1000) / (20 X 1.3) = 196 g/mol Therefore MAB = 196 Let x be the atomic mass of A & y be the atomic mass of B, THEN MAB2 = x + 2y = 110.9 ----------------------------------(1) And MAB = x + 4y = 196        ----------------------------------(2) Subtracting 2 from 1 ,we get 2y  = 196-110.9 y   = 85.1 / 2  y = 42.6 Putting the value of y in 1 we get x = 110.9 - 2 x 42.6 x = 25.59 Therefore atomic mass of A = 25.59 u Atomic mass of B = 42.6 u.
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8 months ago
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