Calculate the molality of KCl solution in water if its freezing point is 271k `( K_f = 1.86k kg mol_(-1) )`

3g urea is dissolved in 100ml water. Find the freezing point of solution ( K_f = 1.86k kg mol_(-1) )

A solution of sucrose (mol.mass=342) was prepared by dissolving 34.2g of it in 1000g water. Find its freezing point. (K_f for water = 1.86k kg mol^(-1) )

A 5% aqueous solution of mass of cane sugar (mw=342) has freezing point 271K. calculate the freezing point of 5% glucose solution in water, if freezing point of pure water is 273.15K.

A solution of urea in water has a boiling point of 100.128^@ C. calculate its freezing point (K_b=0.512 and K_f = 1.86K kg mol_(-1) )

A solution containing 18 g of non - volatile solute in 200g water freeezes as 272.07 K . Calculate the molecular mass of the solute. ( F.P of water = 273 K_f = 1.86 K kg mol^(-1)

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water of the freezing point of pure water is 273.15 K. [Molecular masses glucose C_(6)H_(12)O_(6) = 180 amu, cane sugar C_(12)H_(22)O_(11)= 342 amu]

The freezing point of 1 molar NaCl solution assuming NaCl to be 100% dissociated in water is: ( K_f = 1.86 K molality^-1 )