Verified Solution • 2025–26 Session moderate

QJEEM17Ch87af: JEE Main 2017 Chemistry - Class 12 JEEM17Ch87af Solution: 5 g of Na 2 SO 4 was dissolved in x g of H 2 O. The change in…

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)

Multiple Choice Options

  • A 15 g
  • B 25 g
  • C 45 g
  • D 65 g

Step-by-Step Verified Solution

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  dissociationInitial  no.  of  moles{{Moles\,\,after\,\,dissociation} \over {Initial\,\,no.\,\,of\,\,moles}}

= 1+2x1{{1 + 2x} \over 1}

= 1 + 2 ×\times 0.815

= 2.63

∴   \therefore\,\,\, Δ\Delta Tf = 1000×Kf×w2×iMS×w1{{1000 \times {K_f} \times {w_2} \times i} \over {{M_S} \times {w_1}}}

⇒\Rightarrow    \,\,\, 3.82 = 1000×1.86×2.63×5142×x{{1000 \times 1.86 \times 2.63 \times 5} \over {142 \times x}}

⇒\Rightarrow    \,\,\, x = 45 gm

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