An aqueous solution containing 3,42 grams per 100 grams of water has a freezing point depression of 0,186deg c. What is the molecular mass of the solute?

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Answer 1 · 2018-03-10T03:12:32

#"342 g/mol"#

Freezing point depression is given as

#DeltaT = -K_fm#

Where,

• #DeltaT =# Change in freezing point

• #K_f =# Freezing point constant (It’s #"1.86°C/m"# for water)

• #m =# Molality of solution

#m = (DeltaT)/(-K_f) = (-"0.186 °C")/(-"1.86°C/m") = "0.1 molal"#

#"Molality" = "Moles of solute"/"Mass of solvent (in kg)"#

#"0.1 molal" = "n" / "0.1 kg"#

#"n = 0.01 moles"#

Molar mass of solute #= "w"/"n" = "3.42 g"/"0.01 moles" = color(blue)"342 g/mol"#

That’s molar mass of sucrose.