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Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K.

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Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kg mol-1. Calculate atomic masses of A and B.

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Given,

M2 = $$\frac{1000 \times w_2 \times k_f}{\Delta T_f \times w_1}$$

Then,

MAB2 = $$\frac{1000 \times 1 \times 5.1}{2.3 \times 20}$$

= 110.87 g mol-1

MAB4 = $$\frac{1000 \times 1 \times 5.1}{1.3 \times 20}$$

= 196.15 g mol-1

Now, the molar masses of AB2 and AB4 as 110.87 g mol-1 and 196.15 g mol-1 respectively.

Let the atomic masses of A and B be x and y respectively.

Now, we can write

x + 2y = 110.87   .....(i)

x + 4y = 196.15  .......(ii)

Subtracting equation (i) from (ii), we have

2y = 85.28

⇒ y = 42.64

Putting the value of 'y' in equation (1), we have

x + 2 × 42.64 = 110.87

⇒ x = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

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