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Natalija [7]
3 years ago
15

A solution containing 20.0 g of an unknown liquid and 110.0 g water has a freezing point of .32 °C. Given Kf 1.86°C/m for water,

the molar mass of the unknown liquid is________ g/mol A)256B) 69.0 C) 619 D) 78.1
Chemistry
1 answer:
hjlf3 years ago
4 0

Answer:

A. 256

Explanation:

In a solution where a liquid is the sovent, we'll use the van't Hoff factor, which is the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to 1.

ΔTemp.f = i * Kf * b

where,

ΔTemp.f = the freezing-point depression;

i = the van't Hoff factor

Kf = the cryoscopic constant of the solvent;

b = the molality of the solution.

So the freezing-point depression by definition is the difference between the the freezing point of the pure solvent and the freesing point of the solution.

Mathematically,

ΔTemp.f = Temp.f° - Temp.f

where,

Temp.f° = the freezing point of the pure solvent.

Temp.f = the freezin point of the solution.

Freezing point of pure water = 0°C

ΔTemp.f = 0 - (-1.32)

= 1.32°C

i = 1,

Kf = 1.86 °Ckg/mol

Solving for the molality, b = ΔTemp.f/( i * Kf)

= 1.32/(1*1.86)

= 0.71 mol/kg

Converting from mol/kg to mol/g,

0.71 mol/kg * 1kg/1000g

= 0.00071 mol/g.

Mass of solvent = 110g

Number of moles = mass * molality

= 0.00071 * 110

= 0.078 mol.

To calculate molar mass,

Molar mass (g/mol) = mass/number of moles

Mass of solute (liquid) = 20g

Molar mass = 20/0.078

= 256.2 g/mol

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