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प्रश्न
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.
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उत्तर
We know that,
M2 = `(1000 xx w_2 xx K_f)/(Delta T_f xx w_1)`
Then `M_(AB_2) = (1000 xx 1 xx 5.1)/(2.3 xx 20)`
= 110.87 g mol−1
`M_(AB_4) = (1000 xx 1 xx 5.1)/(1.3 xx 20)`
= 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 + 85.28 = 110.87
x = 110.87 − 85.28
⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u, respectively.
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| S.No | Mass of the salt used in g |
Melting point in °C | |
| Readings Set 1 | Reading Set 2 | ||
| 1 | 0.3 | -1.9 | -1.9 |
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| 6 | 1.0 | -6.4 | -6.3 |
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