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NCERT Class XII Chemistry
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Question : 21
Total: 53
Two elements A and B form compounds having molecular formula AB2 and AB4. When dissolved in 20.0 g of benzene (C6H6), 1.0 g AB2 lowers the freezing point by 2.3°C whereas 1.0 g of AB4 lowers the freezing point by 1.3 °C. The molal depression constant for benzene is 5.1 K kg mol ‌–1. Calculate atomic mass of A and B.
Solution:  
For AB2,ΔTf=Kf×
w2×1000
M2×w1
 or, 2.3=5.1×
1×1000
M2×20
∴‌‌M2=
50×5.1
2.3
=110.86 g mol−1
Similarly for AB4,1.3=5.1×
1×1000
M2×20

∴‌‌M2=
50×5.1
1.3
=196.15 g mol−1
Now, molecular weight of AB2=110.86, molecular weight of AB4=196.15
AB4=A+4B=196.15‌‌‌‌‌‌‌...(i)
AB2=A+2B=110.86‌‌‌‌‌‌‌...(ii)
(i) −( ii) gives 2B=85.29
∴‌‌B=42.645 u
Putting the value of B in equation (ii),
A+2×42.645=110.86 or, A=110.86−85.29=25.57 u
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