A solution made by dissolving 4.71 g of a compound of unknown molar mass in 100.0 g of water has a freezing point of -1.46°C What is the molar mass of the compound?

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Accepted Answer

First substitute the data in [latex]Δt_f=mK_f[/latex] and solve for m:

[latex]Δt_f=+ 1.46 (since the solvent, water, freezes at 0^°C)[/latex]

[latex]K_f=\frac{1.86^°C kg H_2O}{mol solute} [/latex]

[latex]1.46^°C = mK_f = m imes \frac{1.86^°C kg H_2O}{mol solute} [/latex]

[latex]m =\frac{1.46^° \cancel{C} imes mol solute}{1.86^°\cancel{C} imes kg H_2O}=\frac{0.785 mol solute}{kg H_2O}[/latex]

Now convert the data, 4.71 g solute/100.0 g [latex]H_2O[/latex], to g/mol:

[latex](\frac{4.71 g \cancel{solute}}{100.0 \cancel{g H_2O}}) (\frac{1000 \cancel{g H_2O}}{1 \cancel{kg H_2O}} ) (\frac{1 \cancel{kg H_2O}}{0.785 mol \cancel{solute}} )=60.0 g/mol[/latex]

The molar mass of the compound is 60.0 g/mol.

Question 14.16: A solution made by dissolving 4.71 g of a compound of unknow...

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