JEE Main 2017 — Solutions Question with Solution
From: JEE Main 2017 (Online) 8th April Morning Slot
Question
5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be
3.82oC. If Na2SO4 is 81.5% ionised,
the value of x
(Kf for water=1.86oC kg mol−1) is approximately :
(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)
(Kf for water=1.86oC kg mol−1) is approximately :
(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)
Choose an option
Show full solutionCorrect option: C
Correct answer
C45 g
Step-by-step explanation
| Na2SO4 | 2Na+ | + | SO4-2 | ||
|---|---|---|---|---|---|
| Initial | 1 mol | 0 | 0 | ||
| After dissociation | 1 - x | 2x | x |
Total no of Moles after dissociation = 1 + 2x
Na2SO4 is ionised 81.5% means x = 0.815
Von't Hoff factor (i) =
=
= 1 + 2 0.815
= 2.63
Tf =
3.82 =
x = 45 gm
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