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.

Open in App

Solution

Step I: Calculation of molecular masses of compounds $A B_{2}$ and $A B_{4}$
For compound $A B_{2}$
$W_{B} (A B_{2}) = 1 g; W_{A} (C_{6} H_{6}) = 20 g; Δ T_{f} = 2.3 K$
$K_{f} = 5.1 K k g m o l^{- 1} = 5.1 \times 1000 K g m o l^{- 1}$
$M_{B} = \frac{K_{f} \times W_{B}}{Δ T_{f} \times W_{A}} = \frac{(5.1 \times 1000 K g m o l^{- 1}) \times (1 g)}{(2.3 K) \times (20 g)} = 110.87 g m o l^{- 1}$
For compound $A B_{4}$
$W_{B} (A B_{4}) = 1 g; W_{A} (C_{6} H_{6}) = 20 g; Δ T_{f} = 1.3 K K_{f} = 5.1 K k g m o l^{- 1} M_{B} = \frac{(5.1 K k g m o l^{- 1}) \times 1000 \times (1 g)}{(1.3 K) \times (20 g)} = 196.15 g m o l^{- 1}$
Step II: Calculation of the atomic masses of elements A and B
Let the atomic mass of element A = a
Let the atomic mass of element B = b
Molecular mass of $A B_{2} = a + 2 b$
Molecular mass of $A B_{4} = a + 4 b$
According to the available data
a + 2b = 110.87 ..... (i)
a + 4b = 196.15 ..... (ii)
Subtract Eq. (i) from Eq. (ii)
a + 4b - a - 2b = 196.15 - 110.87
$2 b = 85.28; b = \frac{85.28}{2} = 42.64$
Substituting the value of b in Eq. (i)
$a + 2 \times 42.64 = 110.87$
a + 85.28 = 110.87; a = 110.87 - 85.28
= 25.59
This, atomic mass of element A = 25.59
Atomic mass of element B = 42.64

Suggest Corrections

Similar questions

Q. 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

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

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

Q. Two elements