10 gram of solute with molecular mass 100 gram $m o l^{- 1}$ is dissolved in a 100-gram solvent to show ${0.3}^{o} C$ elevation in boiling point. The value of molal ebullioscopic constant will be:

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0.3

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Solution

The correct option is C 0.3
$Δ T_{b} = m \times k_{b}$
$0.3 = \frac{10 \times 1000}{100 \times 100} \times k_{b}$
$k_{b} = 0.3$

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10 gram of solute with molecular mass 100 gram mol−1 is dissolved in a 100-gram solvent to show 0.3oC elevation in boiling point. The value of molal ebullioscopic constant will be:

The correct option is C 0.3ΔTb=m×kb0.3=10×1000100×100×kbkb=0.3

10 gram of solute with molecular mass 100 gram $m o l^{- 1}$ is dissolved in a 100-gram solvent to show ${0.3}^{o} C$ elevation in boiling point. The value of molal ebullioscopic constant will be:

The correct option is C 0.3
$Δ T_{b} = m \times k_{b}$
$0.3 = \frac{10 \times 1000}{100 \times 100} \times k_{b}$
$k_{b} = 0.3$