JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 21)
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)
15 g
25 g
45 g
65 g
Explanation
| Na2SO4 | $$\rightleftharpoons$$ | 2Na+ | + | SO4-2 | |
|---|---|---|---|---|---|
| Initial | 1 mol | 0 | 0 | ||
| After dissociation | 1 - x | 2x | x |
$$\therefore\,\,\,$$ Total no of Moles after dissociation = 1 + 2x
Na2SO4 is ionised 81.5% means x = 0.815
Von't Hoff factor (i) = $${{Moles\,\,after\,\,dissociation} \over {Initial\,\,no.\,\,of\,\,moles}}$$
= $${{1 + 2x} \over 1}$$
= 1 + 2 $$ \times $$ 0.815
= 2.63
$$\therefore\,\,\,$$ $$\Delta $$Tf = $${{1000 \times {K_f} \times {w_2} \times i} \over {{M_S} \times {w_1}}}$$
$$ \Rightarrow $$$$\,\,\,$$ 3.82 = $${{1000 \times 1.86 \times 2.63 \times 5} \over {142 \times x}}$$
$$ \Rightarrow $$$$\,\,\,$$ x = 45 gm
Comments (0)
