Question

What quantity, in grams, of methanol, CH3OH, is required to prepare a 0.244 m solution in 400. g of water?

Answer 1

Answer:3.1213 g of methanol will required to  prepare 0.244 m of solution in 400 g of water.

Explanation;

Let the mass of methanol be x

Mass of the solvent = 400 g= 0,4 kg

1 kg = 1000 g

Molality of solution = 0.244 m

Molality=$\frac{\text{Mass of compound}}{\text{molar mass of compound}\times \text{Weight of the solvent in kg}}$

$0.244 m=\frac{ x }{32 g/mol\times 0.4 kg}$

$x=0.244 mol/kg\times 32 g.mol\times 0.4 kg=3.1232 g$

3.1213 g of methanol will required to  prepare 0.244 m of solution in 400 g of water.

Answer 2

Now. a 1m (that is a 1molal) solution is prepared by disssolving the molar mass in grams of the solute in 1kg or 1000g of solvent 
Therefore a 0.244m solution is prepared by dissolving 32.0420*0.244 = 7.818g CH3OH in 1000g water 
But you want to use only 400g water: 
mass required = 7.818*400/1000 = 3.127g 
Answer: dissolve 3.13g CH3OH in 400g water.