Question

Two elements A and B form compounds having formula \(AB_2\) and \(AB_4\). When dissolved in 20 g of benzene \((C_6H_6)\), 1 g of \(AB_2\) lowers the freezing point by 2.3 K whereas 1.0 g of \(AB_4\) lowers it by 1.3 K. The molar epression constant for benzene is \(5.1 K\, kg mol^{–1}\). Calculate atomic masses of A and B.

Answer 1

We know that,
\(M_2 = \frac{(1000 \times w_2 \times k)}{ΔT_f \times w_1}\)
\(M_{AB_2} = \frac{(1000×1×5.1)}{(2.3×20)}\)
Then,
\(= 110.87 g mol^{-1}\)
\(M_{AB_4} = \frac{(1000×1×5.1)}{(1.3×20)}\)
\(= 196.15 g mol^{-1}\)
Now, we have the molar masses of \(AB_2\) and \(AB_4\) 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
\(x+2 \times 42.64=110.87\)
\(⇒x=25.59\)
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.