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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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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 depression constant for benzene is 5.1 K $kg mol^{–1}$. Calculate atomic masses of A and B.

$\begin{array}{1 1}25.59u,42.64u\\35.59u,12.64u\\45.59u,32.64u\\5.59u,22.64u\end{array} $

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Applying the formula,
$M_2=\large\frac{1000 K_f W_2}{W_1 \times \Delta T_f}$
$M_{AB_2}=\large\frac{1000\times 5.1\times 1}{20\times 2.3}$
$\qquad\;\;=110.87gmol^{-1}$
$M_{AB_4}=\large\frac{1000\times 5.1\times 1}{20\times 1.3}$
$\qquad\;\;=196.15gmol^{-1}$
Suppose atomic masses of A and B are 'a' and 'b' respectively
Then molar mass of $AB_2=a+2b$
$\Rightarrow 110.87gmol^{-1}$
Molar mass of $AB_4=a+4b$
$\Rightarrow 196.15gmol^{-1}$
$\therefore 2b=85.28$
$b=42.64u$
$a+2b=110.87$
$a+2\times 42.64=110.87$
$\Rightarrow a=25.59u$
Atomic mass A=25.59u
Atomic mass B=42.64u
answered Aug 5, 2014 by sreemathi.v
 
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