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3 years ago

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

Chemistry

2 answers:

gladu [14]3 years ago

7 0

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.

siniylev [52]3 years ago

5 0

<span>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.</span>

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2 years ago

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Explanation:

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Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

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A = 191.87 g

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% = (191.87 / 250) * 100

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