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=[tex]\frac{\text{Mass of compound}}{\text{molar mass of compound}\times \text{Weight of the solvent in kg}}[/tex]
[tex]0.244 m=\frac{ x }{32 g/mol\times 0.4 kg}[/tex]
[tex]x=0.244 mol/kg\times 32 g.mol\times 0.4 kg=3.1232 g[/tex]
3.1213 g of methanol will required to prepare 0.244 m of solution in 400 g of water.
The quantity in grams of methanol, is required to prepare a 0.244 m solution in 400. g of water is 3.1232 grams.
Molality is define as the number of moles of solute present in per kilogram of the solvent & it is represented as:
Molality = moles of solute / mass of solvent
Given that mass of solvent = 400g = 0.4kg
molality of mixture = 0.244m
Moles of methanol = (0.244)(0.4) = 0.0976 moles
Now mass from moles will be calculated as:
n = W/M, where
W = given mass of methanol
M = molar mass of methanol = 32g/mol
W = (0.0976)(32) = 3.1232 grams
Hence required mass of methanol is 3.1232 grams.
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