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.