(C) Glycerine is a non-volatile solute. When a non-volatile solute is added to a solvent,it leads to the depression of the freezing point of the solve

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

The freezing point of water is lowered

(C) Glycerine is a non-volatile solute. When a non-volatile solute is added to a solvent,it leads to the depression of the freezing point of the solvent. This is a colligative property known as freezing point depression,represented by the formula $\Delta T_f = K_f \times m$,where $\Delta T_f$ is the depression in freezing point,$K_f$ is the cryoscopic constant,and $m$ is the molality of the solution.

Class 12 Chemistry Solutions Depression of freezing point of the solvent

Medium

When glycerine is added to a litre of water,which of the following behaviors is observed?

A
Water evaporates more easily
B
The temperature of water is increased
C
The freezing point of water is lowered
D
The viscosity of water is lowered

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solutions

Subtopic

Depression of freezing point of the solvent

Which of the following will produce the maximum depression in freezing point of its aqueous solution?

Easy

View Solution

The freezing point of an aqueous solution containing $25 \ g$ of ethanol $(C_2H_5OH)$ in $1000 \ g$ of $H_2O$ is $(K_f = 1.86 \ K \ kg \ mol^{-1})$ (in $^{\circ} C$)

EasyTS EAMCET 2021

View Solution

Calculate the cryoscopic constant $(K_f)$ of a solvent when $2.5 \ g$ of a solute is dissolved in $35 \ g$ of the solvent,which lowers its freezing point by $3 \ K$. (Molar mass of the solute is $117 \ g \ mol^{-1}$)

DifficultMHT CET 2024

View Solution

If the freezing point of a $5\% \ w/w$ aqueous solution of sucrose is $271 \ K$ and pure water has a freezing point of $273.15 \ K$,then calculate the freezing point of a $5\% \ w/w$ aqueous solution of glucose. (in $K$)

Difficult

View Solution

$2.0 \ g$ of a non-electrolyte dissolved in $100 \ g$ of benzene lowers the freezing point of benzene by $1.2 \ K$. The freezing point depression constant of benzene is $5.12 \ K \ kg \ mol^{-1}$. The molar mass of the solute is:

EasyTS EAMCET 2018

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

When glycerine is added to a litre of water,which of the following behaviors is observed?

Similar Questions

Which of the following will produce the maximum depression in freezing point of its aqueous solution?

The freezing point of an aqueous solution containing $25 \ g$ of ethanol $(C_2H_5OH)$ in $1000 \ g$ of $H_2O$ is $(K_f = 1.86 \ K \ kg \ mol^{-1})$ (in $^{\circ} C$)

Calculate the cryoscopic constant $(K_f)$ of a solvent when $2.5 \ g$ of a solute is dissolved in $35 \ g$ of the solvent,which lowers its freezing point by $3 \ K$. (Molar mass of the solute is $117 \ g \ mol^{-1}$)

If the freezing point of a $5\% \ w/w$ aqueous solution of sucrose is $271 \ K$ and pure water has a freezing point of $273.15 \ K$,then calculate the freezing point of a $5\% \ w/w$ aqueous solution of glucose. (in $K$)

$2.0 \ g$ of a non-electrolyte dissolved in $100 \ g$ of benzene lowers the freezing point of benzene by $1.2 \ K$. The freezing point depression constant of benzene is $5.12 \ K \ kg \ mol^{-1}$. The molar mass of the solute is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module