What is the molality (m) of a solution prepared by dissolving 2.70 g CH_3OH in 25.0 g H_2O ?
Question 14.14: What is the molality (m) of a solution prepared by dissolvin...
Since m = \frac{mol solute}{kg solvent} , the conversion is
\frac{2.70 g CH_3OH}{25.0 g H_2O} \rightarrow \frac{mol CH_3OH}{25.0 g H_2O} \rightarrow \frac{mol CH_3OH}{1 kg H_2O}The molar mass of CH_3OH is (12.01 + 4.032 + 16.00), or 32.04 g/mol:
\left(\frac{2.70 \cancel{g CH_3OH}}{25.0 \cancel{g H_2O}}\right) \left(\frac{1 mol CH_3OH}{32.04 \cancel{g CH_3OH}}\right) \left( \frac{1000 \cancel{g H_2O}}{1 kg H_2O}\right) =\frac{3.37 mol CH_3OH}{1 kg H_2O}The molality is 3.37 m.
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