A solution containing 1.8 g of a compound (empirical formula CH 2 O ) in 40 g of water is observed to freeze at - 0 .465 ∘ C . The molecular formula of the compound is- ( K f of water = 1 .86 kg K mol - 1 )

M 2 = K f × w 2 × 1000 △ T f × w 1 or M 2 = 1 .86 × 1 .8 × 1000 0 .465 × 40 = 180 n = 180 emp . formula mass = 180 30 = 6 Molecular formula of the compound is C 6 H 12 O 6 .

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ChemistryLiquid SolutionMiscellaneous/MixedEasy2 minPYQ_2020

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A solution containing 1.8 g of a compound (empirical formula ${CH}_{2} O$ ) in 40 g of water is observed to freeze at $- 0 {.465}^{\circ} C .$ The molecular formula of the compound is-
( $K_{f} of water = 1 .86 kg K {mol}^{- 1}$ )

Options:

Answer:

Solution:

$M_{2} = \frac{K_{f} \times w_{2} \times 1000}{△ T_{f} \times w_{1}}$

$or M_{2} = \frac{1 .86 \times 1 .8 \times 1000}{0 .465 \times 40} = 180$

$n = \frac{180}{emp . formula mass} = \frac{180}{30} = 6$

Molecular formula of the compound is $C_{6} H_{12} O_{6}$ .

Stream:NTA_ABHYASSubject:ChemistryTopic:Liquid SolutionSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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