Elevation in boiling point of a molar (1M) glucose solution (d=1.2 g mL-1) is

Question

Elevation in boiling point of a molar (1M) glucose solution (d=1.2 g mL-1) is (A) 1.34 Kb (B) 0.98 Kb (C) 2.40 Kb (D) Kb

Tardigrade Admin · Accepted Answer

Molarity =1 M=1 mol L-1 1000 mL of solution has glucose = 180 g (1000 × 1.2) g of solution has glucose = 180 g ∴ Solute (w1)=180 g Solution = 1200 g Solvent, H2O (w2)=120-180=1020 g ∴ Δ Tb=(1000Kbw1/m1w2) =(1000× Kb×180/180×1020) =0.98Kb

Q. Elevation in boiling point of a molar (1M) glucose solution $(d = 1.2 g m L^{- 1})$ is

$1.34 K_{b}$

$0.98 K_{b}$

$2.40 K_{b}$

$K_{b}$

Q. Elevation in boiling point of a molar (1M) glucose solution (d=1.2gmL−1) is

1.34Kb​

0.98Kb​

2.40Kb​

Kb​

Molarity =1M=1molL−1 1000 mL of solution has glucose = 180 g (1000×1.2)g of solution has glucose = 180 g ∴ Solute (w1​)=180g Solution =1200g Solvent, H2​O(w2​)=120−180=1020g ∴ΔTb​=m1​w2​1000Kb​w1​​ =180×10201000×Kb​×180​ =0.98Kb​

Q. Elevation in boiling point of a molar (1M) glucose solution $(d = 1.2 g m L^{- 1})$ is

$1.34 K_{b}$

$0.98 K_{b}$

$2.40 K_{b}$

$K_{b}$