Question

A solution is made by dissolving 2.04 g of KBr in 188 g of water. The k_f of water is 1.86 ^oC/m, the k_b of water is 0.51 ^oC/m.

Calculate the freezing point

Calculate the boiling point

Answer 1

delta Tf = Kf * molality

molality = number of moles / mass of solution in kg

number of moles of KBr = 2.04 g / 119.002 g/mol = 0.0171 mole

molality = 0.0171 mole / 0.188 kg = 0.0910 m

delta Tf = 1.86 * 0.0910 = 0.169 ⁰C

Therefore, freezing point = -0.169 ⁰C

delta Tb = Kb * molality

molality = number of moles / mass of solution in kg

number of moles of KBr = 2.04 g / 119.002 g/mol = 0.0171 mole

molality = 0.0171 mole / 0.188 kg = 0.0910 m

delta Tb = 0.51 * 0.0910 = 0.0464 ⁰C

Therefore, boiling point = 100.0464 ⁰C