Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol−1. Calculate atomic masses of A and B.
We know that,
`M_2 = (1000xxw_2xxk_f)/(triangleT_fxxw_1)`
Then `M_(AB_2) = (1000xx1xx5.1)/(2.3 xx 20)`
= 110.87 gmol-1
`M_(AB_4) = (1000xx 1xx5.1)/(1.3xx20)`
= 196.15 g mol−1
Now, we have the molar masses of AB2 and AB4 as 110.87 g mol−1 and 196.15 g mol−1 respectively.
Let the atomic masses of A and B be x and y respectively.
Now, we can write:
x + 2y = 110.87 (i)
x +4y = 196.15 (ii)
Subtracting equation (i) from (ii), we have
2y = 85.28
⇒ y = 42.64
Putting the value of ‘y’ in equation (1), we have
x + 2 × 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.
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