What is the molality of a D-glucose solution prepared by dissolving 36.0 g of D-glucose, C6H12O6, in 125.9 g of water

Chemistry

1 answer:

mestny [16]2 years ago

6 0

Answer:

1.59 molal

Explanation:

The definition of molality is:

molality = moles solute / kg of solvent

So first we calculate the moles of solute (D-glucose, C₆H₁₂O₆), using its molar mass:

36.0 g C₆H₁₂O₆ ÷ 180.156 g/mol = 0.200 mol C₆H₁₂O₆

Then we convert grams of water into kg:

125.9 g H₂O * $\frac{1kg}{1000g}$ = 0.1259 kg H₂O

Finally we calculate the molality:

m = 0.200 mol / 0.1259 kg

m = 1.59 m

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