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Solutions

JEE Main level, numerical answer, medium

2 of 4

Question 2 of 4

The elevation in boiling point when $1\,\text{mole}$ of non-volatile solute is dissolved in $1000\,\text{g}$ of solvent ( $K_b = 0.52\,\text{K}\cdot\text{kg/mol}$ ) is (in K):

Show answer and solution

Answer: 0.52

\Delta T_b = K_b \times m = 0.52 \times 1.0 = 0.52\,\text{K}

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The elevation in boiling point when 1 mole1\,\text{mole}1mole of non-volatile solute is dissolved in 1000 g1000\,\text{g}1000g of solvent (Kb=0.52 K⋅kg/molK_b = 0.52\,\text{K}\cdot\text{kg/mol}Kb​=0.52K⋅kg/mol) is (in K):

JEE Main question paper PDF

More Chemistry practice

The elevation in boiling point when $1\,\text{mole}$ of non-volatile solute is dissolved in $1000\,\text{g}$ of solvent ( $K_b = 0.52\,\text{K}\cdot\text{kg/mol}$ ) is (in K):