SATHEE: Solutions Question 347

Solutions Question 347

Question: A solution containing 0.85 g of $Z n C l_{2}$ in 125.0 g of water freezes at $- 0.23^{\circ} C$ . The apparent degree of dissociation of the salt is ( $K_{f}$ for water $= 1.86 K k g m o l^{- 1}$ , atomic mass: Zn = 65.3 and Cl = 35.5)

Options:

A) 1.36%

B) 73.5%

C) 7.35%

D) 2.47%

Show Answer

Answer:

Correct Answer: B

Solution:

$M o l . w t . = \frac{K_{f} \times w \times 1000}{Δ T_{f} \times W}$

$= \frac{1.86 \times 0.85 \times 1000}{0.23 \times 125} \approx 55 g$ Where $w = 0.85 g$

$W = 125 g$

$Δ T_{f} = 0^{\circ} C - (- 23^{\circ} C) = 23^{\circ} C$ Now, $i = \frac{M_{n o r m a l}}{M_{o b s e r v e d}} = \frac{136.3}{55} = 2.47$

$\underset{1 - α}{Z n C l_{2}}, \to \underset{α}{Z n},^{+ +} + 2 \underset{2 α}{C l^{-}},$ Van’t Hoff factor (i) $= \frac{1 - α + α + 2 α}{1} = 2.47$
$∴, α = 0.735 = 73.5$

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Solutions Question 347

Question: A solution containing 0.85 g of ZnCl2 in 125.0 g of water freezes at −0.23∘C . The apparent degree of dissociation of the salt is ( Kf for water =1.86Kkgmol−1 , atomic mass: Zn = 65.3 and Cl = 35.5)

Options:

Answer:

Solution:

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Question: A solution containing 0.85 g of $Z n C l_{2}$ in 125.0 g of water freezes at $- 0.23^{\circ} C$ . The apparent degree of dissociation of the salt is ( $K_{f}$ for water $= 1.86 K k g m o l^{- 1}$ , atomic mass: Zn = 65.3 and Cl = 35.5)