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

Open in App

Solution

In case of compound $A B_{2}$ :

$M_{B} = \frac{K_{f} W_{B} \times 1000}{W_{A} Δ T_{f}}$
$Δ T_{f} = 2.3 K, W_{B} = 1.0 g, W_{A} = 20.0 g, K_{f} = 5.1 K K g / m o l$

$M_{B} = \frac{5.1 \times 1.0 \times 1000}{20.0 \times 2.3} = 110.87$ g/mol
In case of compound $A B_{4}$ :

$Δ T_{f} = 1.3 K, W_{B} = 1.0 g, W_{A} = 20.0 g$
$M_{B} = \frac{5.1 \times 1.0 \times 1000}{20.0 \times 1.3} = 196.15$ g/mol
Let $a$ g/mol and $b$ g/mol be the atomic masses of $A$ and $B$ respectively.

$M_{A B_{2}} = a + 2 b = 110.87$ $. . . . . (i)$

$M_{A B_{4}} = a + 4 b = 196.15$ $. . . . . . . (i i)$

Substracting equation $(i i)$ from equation $(i)$ , we have

$- 2 b = - 85.28$

Atomic mass of B is $b = 42.64$ .

Substituting the values of $b$ in equation $(i)$ , we get

$a + 2 \times 42.64 = 110.87$

Atomic mass of A is $a = 25.59$ g/mol.

Hence, the atomic mass of a is 25.59 g/mol and b is 42.64 g/mol

Suggest Corrections

Similar questions

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.

Two elements A and B form compounds of formula AB₂ and AB₄. When dissolved in 20.0g of benzene 1.0g of AB₂lowers freezing point by 2.3^°C whereas 1.0g of AB₄ lowers freezing point by 1.3^°C. The K_f for benzene. The atomic masses of A and B are