When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH, CN), the freezing point of the solution is measured to be 13.4 °C. Calculate the molar mass of X.

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)

ΔT = 13.4°C - (-12.82) = 26.22°C

m is molality of the solution

Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)

And i is Van't Hoff factor (1 for all solutes in benzonitrile)

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

0.551g / 0.490mol

= 1.12g/mol

This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.

if the temperature drops and becomes cool

then the atmosphere can be freezing as the water vapor in the air is at maximum (100%) and can be severely cold

also can be snowing lightly as the moisture in the air will start to solidify

and btw this is not chemistry but