Answer to Question #299520 in General Chemistry for nav

Question

1. How many grams of Lactic acid (C₆H₁₀O₅) should be dissolved in 650g of cyclohexane (C₆H₁₂) to raise the boiling point to 84.91⁰C? The normal boiling point of cyclohexane is 81⁰C and its molal boiling point is 2.75 ⁰C/m.

Answer 1

!!! C₆H₁₀O₅ is lactic acid lactate because lactic acid is C₃H₆O₃ !!!

Solution:

Δt = i × K_b × m

where:

Δt = the change in boiling point

i = Vant Hoff's coefficient

K_b = the molal boiling-point elevation constant

m = the molality of the solute

Δt = 84.91°C - 81°C = 3.91°C

i = 1 (for lactic acid lactate - nonelectrolyte)

K_b = 2.75°C/m (for cyclohexane)

Therefore,

m = Δt / (i × K_b)

m = (3.91°C) / (1 × 2.75°C/m) = 1.422 m

Molality of C₆H₁₀O₅ = 1.422 m

Molality of C₆H₁₀O₅ = Moles of C₆H₁₀O₅ / Kilograms of cyclohexane

Therefore,

Moles of C₆H₁₀O₅ = Molality of C₆H₁₀O₅ × Kilograms of cyclohexane

Moles of C₆H₁₀O₅ = (1.422 m) × (0.65 kg) = 0.9243 mol

The molar mass of lactic acid lactate (C₆H₁₀O₅) is 162.14 g/mol

Therefore,

Mass of C₆H₁₀O₅ = (0.9243 mol) × (162.14 g / 1 mol) = 149.87 g

Mass of C₆H₁₀O₅ = 149.87 g

Answer: 149.87 grams of C₆H₁₀O₅ should be dissolved.