How many gram of sucrose (mol. wt. = 342) should be dissolved in 100 g water in order to produce a solution with a 105.0° C difference between the freezing point and boiling temperature? [kf=1.860° C text/m, kb=0.151° C text/m]

Question

How many gram of sucrose (mol. wt. = 342) should be dissolved in 100 g water in order to produce a solution with a 105.0° C difference between the freezing point and boiling temperature? [kf=1.860° C text/m, kb=0.151° C text/m]  (A) 34.2 g (B) 72 g (C) 342 g (D) 460 g

Tardigrade Admin · Accepted Answer

Boiling point (Tb)=100+Δ Tb=100+kbm Freezing point (Tf)=0-Δ Tf=-kfm Tb-Tf=(100+kbm)-(-kfm) 105=100+0.51 m+1.86 m 2.37 m=5 or m=(5/2.37) =2.11 ∴ Weight of sucrose to be dissolved in 100g water =(2.11× 342/1000)× 100 =72 g

Q. How many gram of sucrose (mol. wt. = 342) should be dissolved in 100 g water in order to produce a solution with a 105.0∘C difference between the freezing point and boiling temperature? [kf​=1.860∘C/m,kb​=0.151∘C/m]

34.2 g

72 g

342 g

460 g

Boiling point (Tb​)=100+ΔTb​=100+kb​m Freezing point (Tf​)=0−ΔTf​=−kf​m Tb​−Tf​=(100+kb​m)−(−kf​m) 105=100+0.51m+1.86m 2.37m=5 or m=2.375​ =2.11 ∴ Weight of sucrose to be dissolved in 100g water =10002.11×342​×100 =72g

Q. How many gram of sucrose (mol. wt. = 342) should be dissolved in 100 g water in order to produce a solution with a $105.0^{\circ} C$ difference between the freezing point and boiling temperature? $[k_{f} = 1.860^{\circ} C / m, k_{b} = 0.151^{\circ} C / m]$