Question

3) Determine the temperature change of 25.6 grams of HOCH₂CH₂OH dissolved in 475 ml of water. The k_f = 1.86 C / Molal. Will an equal amount of CH₃OH give a bigger or smaller temperature change? Explain

Answer 1

First, the equation of freezing point depression

dTf = -Kf*m

where m = molality = mol of solute / kg of solvent

mass of water = 475 mL --> 475 g = 0.475 kg

Kf = -1.86 C/m is constant

then

mol of solute = mass/MW = (25.6)/(MW)

MW for EG = 62,0678 G/mol

MW of methanol = 32.06

Substitute data

dTf = -1.86 *(25.6/MW ) /(0.475)

Substitute

a) Ethyleneglycol --> dTf = -1.86 *(25.6/62.02) /(0.475) = -1.616 °C

b) MEthanol --> dTf = -1.86 *(25.6/32.6) /(0.475) = -3.074 °C

Cearly, the methanol mix will be LOWER, since it has much more moles per unit mas... therefore, decreases more