The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is 0.1° C. The molal elevation constant of the liquid is

Question

The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is 0.1° C. The molal elevation constant of the liquid is (A) 2 K kg / mol (B) 10 K kg / mol (C) 0.1 K kg / mol (D) 1 K kg / mol

Tardigrade Admin · Accepted Answer

Δ T b = K b . m . i ⇒ 0.1= K b × (1.8/180) × (1000/100) × 1 K b =1

Q. The rise in boiling point of a solution containing $1.8 g$ of glucose in $100 g$ of solvent is $0. 1^{\circ} C$ . The molal elevation constant of the liquid is

$2 K k g / m o l$

$10 K k g / m o l$

$0.1 K k g / m o l$

$1 K k g / m o l$

$Δ T_{b} = K_{b} . m . i$
$\Rightarrow 0.1 = K_{b} \times \frac{1.8}{180} \times \frac{1000}{100} \times 1$
$K_{b} = 1$

Q. The rise in boiling point of a solution containing 1.8g of glucose in 100g of solvent is 0.1∘C. The molal elevation constant of the liquid is

2Kkg/mol

10Kkg/mol

0.1Kkg/mol

1Kkg/mol