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Byju's Answer

Standard XII

Chemistry

Elevation in Boiling Point

A solution co...

Question

A solution containing 7 g of a solute (molar mass 210 g $m o l^{- 1}$ ) in 350 g of acetone raised the boiling point of acetone from $56^{o} C$ to ${56.3}^{o} C$ . The value of ebullioscopic constant of acetone in K kg $m o l^{- 1}$ is :

2.66

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3.15

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4.12

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2.86

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Solution

The correct option is A 3.15
we have molality = $\frac{n o . o f m o l e s o f s o l u t e}{w e i g h t o f s o l v e n t}$
$Δ T_{b} = K_{b} * m$
$56.3 - 56 = K_{b} * \frac{7 * 1000}{210 * 350}$
we get ebullioscopic constant = 3.15 $K K g m o l^{- 1}$

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A solution containing 7 g of a solute (molar mass 210 g mol−1) in 350 g of acetone raised the boiling point of acetone from 56oC to 56.3oC. The value of ebullioscopic constant of acetone in K kg mol−1 is :

The correct option is A 3.15we have molality = no.ofmolesofsoluteweightofsolventΔTb=Kb∗m56.3−56=Kb∗7∗1000210∗350we get ebullioscopic constant = 3.15 KKgmol−1

A solution containing 7 g of a solute (molar mass 210 g $m o l^{- 1}$ ) in 350 g of acetone raised the boiling point of acetone from $56^{o} C$ to ${56.3}^{o} C$ . The value of ebullioscopic constant of acetone in K kg $m o l^{- 1}$ is :

The correct option is A 3.15
we have molality = $\frac{n o . o f m o l e s o f s o l u t e}{w e i g h t o f s o l v e n t}$
$Δ T_{b} = K_{b} * m$
$56.3 - 56 = K_{b} * \frac{7 * 1000}{210 * 350}$
we get ebullioscopic constant = 3.15 $K K g m o l^{- 1}$