Question

Methanol (CH₃OH, MW = 32) is a nonelectrolyte. What is the amount (in grams) of methanol that must be added to 600 ml of water to produce a boiling point elevation of 0.15^OC? The K_b value for water is 0.51 (degree-kg/mole).

Answer 1

ANSWER:

Given,

Molecular weight solute = 32 g/mol

Molecular weight of solvent = 18 g/mol

Volume of solvent = 600 mL

Density of water = 1 g/mL

T_f = 0.15 ^oC

K_f = 0.51 ^oC-kg/mole

Now,

Molality= (weight of solute in gram x 1000)/(mol. wt. of solute x weight of solvent in grams)

Molality (m) = (weight of solute in gram x 1000)/(32 g/mol x 600 g)

= (0.0521 x weight of solute in gram) mol/kg

And,

T_f = K_f x molality

0.15 ^oC = 0.51 ^oC-kg/mole x (0.0521 x weight of solute in gram)

Weight of solute = 5.65 g

Hence, weight of methanol is 5.65 gram, which is added to 600 mL of water to produce a boiling point elevation of 0.15 ^oC.