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प्रश्न
When 25.6 g of sulphur was dissolved in 1000 g of benzene, the freezing point lowered by 0.512 K. Calculate the formula of sulphur (Sr).
(Kf for benzene = 5.12 K kg mol−1, Atomic mass of sulphur = 32 g mol−1)
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उत्तर
Kf for benzene = 5.12 K kg mol−1
WB = mass of solute = 25.6 g
WA = 1000 g
ΔTf = 5.12 K kg mol−1
Atomic mass of sulphur = 32 g mol−1
MB = `("K"_"f" xx "W"_"B" xx 1000)/(Δ"T"_"f" xx "W"_"A")`
= `(5.12 xx 25.6 xx 1000)/(0.512 xx 1000)`
= `131.072/0.152`
= 256
Now the molecular mass of
Sx = x × 32 = 256
x = `256/32` = 8
Therefore, the formula of sulphur = S8.
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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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