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Standard XII

Chemistry

Depression in Freezing Point

Two elements ...

Question

Two elements $A$ and $B$ form compounds having molecular formulae $A B_{2}$ and $A B_{4} .$ When dissolved in $20.0$ g of benzene, $1.0$ g of $A B_{2}$ lowers the freezing point by ${2.3}^{o} C$ , whereas $1.0$ g of $A B_{4}$ lowers the freezing point by ${1.3}^{o} C$ . The molal depression constant for benzene in $1000$ g is $5.1$ . Calculate the atomic masses of $A$ and $B$ .

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Solution

We know that,
$m = \frac{1000 K_{f} \times w}{W \times △ T}$
Molecular mass of $A B_{2}$ (from given data) = $\frac{1000 \times 5.1 \times 1}{20 \times 2.3}$
$= 110.86$
and Molecular mass of $A B_{4}$ (from given data) $= \frac{100 \times 5.1 \times 1}{1.3 \times 20}$
$= 196.15$
Further, $A B_{4} = A + 4 B = 196.15$ ...(i)
$A B_{2} = A + 2 B = 110.86$ ...(ii)
Subtracting eq. (ii) from (i),
$2 B = 85.29$
$B = 42.645$
Putting the value of B in eq. (ii),
$A + 85.29 = 110.86$
or $A = 110.86 - 85.29 = 25.57$
Thus, the atomic masses of $A$ and $B$ are $25.57$ and $42.645$ respectively.

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QUESTION 2.21

Two elements A and B form compounds having formula $A B_{2}$ and $A B_{4}$ . When dissolved in 20 g of benzene $(C_{6} H_{6})$ , 1 g $A B_{2}$ lowers the freezing point by 2.3 K whereas 1.0 g of $A B_{4}$ lowers it by 1.3K. The molal depression constant for benzene is 5.1 K kg $m o l^{- 1}$ . Calculate atomic masses of A and B.