Question

A 100.0 mL aqueous sodium chloride solution is 13.9 % NaCl by mass and has a density of 1.12 g mL−1.

What mass of solute would you add to make the boiling point of the solution 104.8  ∘C ? (Use i = 1.8 for NaCl)

Answer 1

delta T = 104.8 - 100 ( pure water boiling point )

= 4.8 C

again,

delta T = i*Kb*molality

(Kb = 0.512*C/m) & i=1.8 ( Na+ & Cl- )

= 4.8 = 1.8*0.512*molality

= molality = 5.2084

now, for solution

mass of solution = volume * density

= 100 * 1.12 = 112 grams

mass of NaCl = mass percent * total mass of the solution / 100

= 13.9* 112/100

= 15.568

mass of solvent = mass of solution - mass of solute

= 112 - 15.568

= 96.432 grams or 0.096432 Kg

moles of NaCl = mass / molar mass

= 15.568 / 58.5

= 0.266

molality = moles / mass of solvent(in kg)

= 0.266/ 0.096432

= 2.76