A 5% solution (by mass) of cane sugar in water has freezing point of 27IK Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

A 0.1539 molal aqueous solution of cane sugar ( molar mass-342gmol^-1 ) has a freezing of 271K while the freezing point of pure water is 273.15K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol.mass=180gmol^-1) per 100 g of solution.

The freezing point of equimolar aqueous solutions will be highest for

Which of the following aqueous solution has the higher freezing point?

Define the terms given below and answer the questions associated with them.Freezing point: What is the freezing point of water?

A 5% solution of cane sugar (molar mass=342) is isotonic with 0.877% soluton of urea. Calculate the molar mass of urea.

Freezing point of a solvent containing a non volatile solute

Calculate the amount of KCl which must be added to 1 kg of water so that its freezing point is depressed by 2 K.

What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50^@C ? The freezing point depression constant (k_f) for water is 1.86^@C/m . Assume that Van't Hoff factor for NaCl is 1.87. (Molar mass of NaCl =58.5g).

What mass of NaCl (molar mass= 58.5 g mol^-1 ) must be dissolved in 65.0 g of water to lower the freezing point by 7.50 ^@C ? The freezing point depression constant, K_f , for water is 1.86 K kg mol^-1 . Assume Van's Hoff factor for NaCl is 1.87.

True or false The depression in freezing point for 1 m solution of a solute in water and benzene is same.