A student wants to prepare a 0.050M solution of NaBr in water. How many grams of

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? 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?
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?

Explanation / Answer

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

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