5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82°C. If Na2SO4 is 81.5 % ionised, the value of x (Kf for water = 1.86°C kg mol-1) is approximately: (molar mass of S = 32 g mol-1 and that of Na = 23 g mol-1)

Question

5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82°C. If Na2SO4 is 81.5 % ionised, the value of x (Kf for water = 1.86°C kg mol-1) is approximately: (molar mass of S = 32 g mol-1 and that of Na = 23 g mol-1) (A) 15 g (B) 25 g (C) 45 g (D) 65 g

Tardigrade Admin · Accepted Answer

Na2SO4 arrow 2Na++SO42- x = 1 + (3 - 1) 0.815 = 2.63 3.82 = 1.86 Ã 2.63×(5× 1000/142× x) ∴ x=(1.86× 2.63× 5000/142× 3.82) =45 gm

$15 g$

$25 g$

$45 g$

$65 g$

$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{2 -}$
$x = 1 + (3 - 1) 0.815 = 2.63$
$3.82 = 1.86 \times 2.63 \times \frac{5 \times 1000}{142 \times x}$
$∴ x = \frac{1.86 \times 2.63 \times 5000}{142 \times 3.82}$
$= 45 g m$

Q. 5g of Na2​SO4​ was dissolved in xg of H2​O. The change in freezing point was found to be 3.82∘C. If Na2​SO4​ is 81.5% ionised, the value of x (Kf​ for water = 1.86∘Ckgmol−1) is approximately : (molar mass of S=32gmol−1 and that of Na=23gmol−1)

15g

25g

45g

65g

Na2​SO4​→2Na++SO42−​ x=1+(3−1)0.815=2.63 3.82=1.86×2.63×142×x5×1000​ ∴x=142×3.821.86×2.63×5000​ =45gm

$15 g$

$25 g$

$45 g$

$65 g$

$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{2 -}$
$x = 1 + (3 - 1) 0.815 = 2.63$
$3.82 = 1.86 \times 2.63 \times \frac{5 \times 1000}{142 \times x}$
$∴ x = \frac{1.86 \times 2.63 \times 5000}{142 \times 3.82}$
$= 45 g m$