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Solutions — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistrySolutions3 Marks
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 depression constant for benzene is $5.1\ K\ kg\ mol^{–1}$. Calculate atomic masses of $A$ and $B.$
✓

Answer

$\text{Using the relation},\text{ M}_2=\frac{1000\times\text{k}_\text{f}\times\text{w}_2}{\text{w}_1\times\Delta\text{T}_\text{f}}$$\therefore\ \text{M}_{\text{AB}_2}=\frac{1000\times5.1\times1}{20\times2.3}=110.87\text{ g mol}^{-1}$
$\text{M}_{\text{AB}_4}=\frac{1000\times5.1\times1}{20\times1.3}=196.15\text{ g mol}^{-1}$
Let the atomic masses of A and B are 'p' and 'q' respectively.
Then molar mass of
$AB_2= p + 2q = 110.87\ g\ mol^{-1} .... (i)$
And molar mass of
$AB_4 = p + 4q = 196.15\ g\ mol^{-1}.... (ii)$
Substracting equation (ii) from equation(i), we
get $2q = 85.28 ⇒ q = 42.64$
Putting $q = 42.64$ in equ. (i), we get
$p = 110.87 - 85.28$
$p = 25.59$
Thus, atomic mass of $A = 25.59\ g\ mol^{-1}$ and atomic mass of $B = 42.64\ g\ mol^{-1}$

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