A. 0.2436 sample of an unknown substance was dissolved in 20.0mL of cyclohexane. The density of cyclohexane is 0.779 g/mL. The freezing point depression was 2.50 oC and the Kf value for cyclohecane is 20.5oC/m. Calculate the molality of the above solution and the molar mass of unknown.

The molality of solution is 0.122 m and molar mass of unknown substance is 128.16 g/mol

Explanation:

where,

= depression in freezing point = 2.50°C

i = Vant hoff factor = 1 (For non-electrolytes)

= molal freezing point elevation constant = 20.5°C/m

m = molality of solution = ?

Putting values in above equation, we get:

Density of cyclohexane = 0.779 g/mL

Volume of cyclohexane = 20.0 mL

Putting values in above equation, we get:

where,

= Given mass of solute = 0.2436 g

= Molar mass of solute = ? g/mol

= Mass of solvent (cyclohexane) = 15.58 g

Molality of solution = 0.122 m

Putting values in above equation, we get:

Hence, the molality of solution is 0.122 m and molar mass of unknown substance is 128.16 g/mol

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

confused

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