Question

A student wants to prepare a 0.050M solution of NaBr in water. How many grams of NaBr are needed to make 500mL of the solution? Given the molar mass of sodium bromide = 102.9g/mol.

What is the molality of this solution?

What is the freezing point of this solution?

Given Kf= 1.86 degrees Celcius/m for water?

What is the boiling point of this solution?

Given Kf= 0.512 degrees Celcius/m for water?

What is the osmotic pressure of this solution at 25 degrees Celcius?

Answer 1

SInce molar mass = 102.9g/mol

Mass of 0.05 mole = 102.9*0.05 = 5.145 g

Mass of 0.025 mole = 2.5725 g (for one liter it is 5.145 gram so it will be half for 500 mL)

In this case molality = moles of solute / mass of solvent in Kg

Mass of solute = 500-2.5725 = 497.4275 g

= 0.4974 Kg

Molality m = 0.025/0.4974

= 0.05026

dTf = Kf*m*1

(0-T) = 1.86*0.05026

Tf = -1.86 'C

Elevation in boiling point = dTb = Kb*m*i (( here i = 2)

dTb = 0.512*0.05026*2 = 0.050

Boiling point = 100.050 'C