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Question
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 K kg mol−1. Calculate the atomic masses of A and B.
Solution
We know that,
M2 = `(1000xx"w"_2xx"K"_"f")/(triangle"T"_"f"xx"w"_1)`
Then `"M"_("AB"_2) = (1000xx1xx5.1)/(2.3 xx 20)`
= 110.87 g mol−1
`"M"_("AB"_4) = (1000xx 1xx5.1)/(1.3xx20)`
= 196.15 g mol−1
Now, the molar masses of AB2 and AB4 are 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 (i), 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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