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 Kwhereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol-1. Calculate atomic masses of A and B.
As We know that:
MB = (Kf x wB x 1000) / (wA X ΔTf)
Now ΔTf = 2.3 , wB = 1.0 , wA = 20, KF = 5.1 (given)
PUTTING THE VALUES IN THE EQUATION
MB = (5.1 x 1 x 1000) / (20 x 2.3) = 110.87 g/mol
Therefore MAB2 = 110.9
For AB4 compound
ΔTf = 1.3 , wb = 1 ,wa = 20
MB = (5.1 X 1 X 1000) / (20 X 1.3) = 196 g/mol
Therefore MAB = 196
Let x be the atomic mass of A & y be the atomic mass of B,
THEN MAB2 = x + 2y = 110.9 ----------------------------------(1)
And MAB = x + 4y = 196 ----------------------------------(2)
Subtracting 2 from 1 ,we get
2y = 196-110.9
y = 85.1 / 2
y = 42.6
Putting the value of y in 1 we get
x = 110.9 - 2 x 42.6
x = 25.7
Therefore atomic mass of A = 25.7 u Atomic mass of B = 42.6 u.
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3 Comment(s) on this Question
I Like this solutions
In this question we are calculating the molecular mass of AB2 and AB4 molecule separately,therefore in case molecular mass of AB2 is taken and in other case molecular mass of AB4 is taken
in molality we take the molecular mass of solution in denominator but here there is only molecular mass of compound AB2.Why?
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