Question

The rise in the boiling point of a solution containing 1.8 g of glucose in 100 g. of a solvent is $0.1^{\circ }C.$ The molal elevation constant of the liquid is ___________.

• A. ${0.021}^{o}C$
• B. ${0.1}^{o}C$
• C. $1^{o}C$
• D. ${10}^{o}C$

Answer 1

The correct option is C $1^{o}C$

$\text{Elevation in boiling point:}$

$T_b-T_b^o=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

$where,$

$T_b=\text{boiling point of solution}$

$T_b^o=\text{boiling point of solvent}$

$k_b=\text{molar elevation constant = ?}$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte such as glucose )

$w_2=\text{mass of solute (glucose) = 1.8 g}$

$w_1=\text{mass of solvent = 100 g}$

$M_2=\text{molar mass of solute (glucose) = 180g/mol}$

$\text{Now put all the given values in the above formula, we get:}$

$(0.1{)}^{o}C=1\times k_b\times \frac{\left(1.8g\right)\times 1000}{180\times 100g}$

$k_b={0.1}^o$

$\text{Therefore, the molar elevation of constant of liquid is}{0.1}^{o}C$