SATHEE: Solutions Question 354

Solutions Question 354

Question: A solution containing 1.8 g of a compound (empirical formula $C H_{2} O$ ) in 40 g of water is observed to freeze at $- 0.465^{\circ} C$ . The molecular formula of the compound is ( $K_{f}$ of water $= 1.86 k g K m o l^{- 1}$ )

Options:

A) $C_{2} H_{4} O_{2}$

B) $C_{3} H_{6} O_{3}$

C) $C_{4} H_{8} O_{4}$

D) $C_{6} H_{12} O_{6}$

Show Answer

Answer:

Correct Answer: D

Solution:

$Δ T_{f} = K_{f} \times m$

$M = \frac{1000 \times K_{f} \times w_{2} (s o l u t e)}{Δ T_{f} \times w_{1} (s o l v e n t)}$

$= \frac{1000 \times 1.86 \times 1.8}{0.465 \times 40} \Rightarrow M = 180$ Molecular formula = ${(e m p i r i c a l f o r m u l a e)}_{n}$

$n = \frac{M o l e c u l a r m a s s}{E m p i r i c a l f o r m u l a m a s s} = \frac{180 g}{30} = 6$ Molecular formula $= {(C H_{2} O)}_{6} = C_{6} H_{12} O_{6}$

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Solutions Question 354

Question: A solution containing 1.8 g of a compound (empirical formula CH2O ) in 40 g of water is observed to freeze at −0.465∘C . The molecular formula of the compound is ( Kf of water =1.86kgKmol−1 )

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Answer:

Solution:

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Question: A solution containing 1.8 g of a compound (empirical formula $C H_{2} O$ ) in 40 g of water is observed to freeze at $- 0.465^{\circ} C$ . The molecular formula of the compound is ( $K_{f}$ of water $= 1.86 k g K m o l^{- 1}$ )