Question

aqueous solution containing 17.5 g of an unknown nonelectrolyte compound in 100. gof water has a freezing point of –1.8 oC. Calculate the molar mass of the unknown compound. Kffor water = 1.86 °C/m

Answer 1

Answer:

The molar mass of the nonelektrolyte is 180.8 g/mol

Explanation:

Step 1: Data given

Mass of non-elektrolyte = 17.5 grams

MAss of water = 100 grams = 0.1 kg

Freezing point of solution = -1.8 °C

Kf water = 1.86 °C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ with ΔT = the freezing point depression = 1.8 °C

⇒ with i = the van't Hoff factor = 1

⇒ with kf = the freezing point depression constant = 1.86 °C/m

⇒ with m = molality

1.8 = 1* 1.86 * m

m =1.8 / 1.86

m = 0.9677 molal

Step 3: Calculate moles nonelectrolyte

molality = moles nonelectrolyte / mass water

0.9677 molal = moles / 0.1 kg

moles = 0.9677 molal * 0.1 kg

moles = 0.09677 moles

Step 4: calculate molar mass of the nonelectrolyte

Molar mass = mass / moles

Molar mass = 17.5 grams / 0.09677 moles

Molar mass = 180.8 g/mol

The molar mass of the nonelektrolyte is 180.8 g/mol

Answer 2

Answer:

The molar mass for the solute is 180.8 g/mol

Explanation:

We apply the colligative property of freezing point depression.

The formula is T° freezing pure solvent - T° freezing solution = Kf . m

T° freezing pure solvent - T° freezing solution → ΔT

Kf = 1.86 °C/m

m is molality. We must determine with data given.

Molality are the moles of solute contained in 1 kg of solvent

0° - (-1.8°C) = 1.86 °C/m . m

1.8°C / 1.86 m/°C = m → 0.968 mol/kg

We multiply the molality by kg of solute, to determine the moles

100 g . 1kg/1000 g = 0.1kg → 0.968 mol/kg . 0.1 kg = 0.0968 moles

To find the molar mass we make g/ mol → 17.5 g / 0.0968 mol =

180.8 g/mol