Kf(water) = -1.86 °C/m In a laboratory experiment, students synthesized a new compound and found that when 13.39 grams of the compound were dissolved in 202.1 grams of water, the solution began to freeze at -2.050 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?

The molecular weight for the compound is 60.1 g/mol

Explanation:

We need to determine the molality of solute to find out the molar mass of it.

We apply the colligative property of freezing point depression:

ΔT = Kf . m . i

If the compound was also found to be nonvolatile and a non-electrolyte,

i = 1.

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

0°C - (-2.05°C) = 1.86°C/m . m

2.05°C / 1.86m/°C = m → 1.10 mol/kg

To determine the moles of solute we used, we can multiply molality by the mass of solvent in kg → 202.1 g . 1kg/1000g = 0.2021 kg

1.10 mol/kg . 0.2021kg = 0.223 moles

Molar mass→ g/mol → 13.39 g / 0.223 mol = 60.1 g/mol

it should be p1 v1 = p2 v2

law of conservation of matter