What is the molality of a d-mannose solution prepared by dissolving 45.0 g of d-mannose, c6h12o6, in 125.9 g of water?

Molality is a unit of concentration that represents the moles of solute per kilogram of solvent. The solvent here is water.

Mass of solvent in kg = 125.9 g * 1 kg/1000 g = 0.1259 kg

The moles of d-mannose knowing that its molar mass is 180.156 g/mol,

Moles solute = 45 g * 1 mol/180.156 g = 0.2498 mol

Thus,
Molality = 0.2498/0.1259 = 1.98 m

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tutorareej tutorareej · Answer 2 · 2021-10-12T11:03:27+02:00

Answer:

The molality of 45 g of D- mannose in 125.9 g of water will be 1.985 m.

Explanation:

Molality is the number of moles of a solute in a solution per kg of solvent.

Molality = [tex]\rm \frac{weight}{molecular weight}\;\times\;\frac{1}{mass\;(kg)}[/tex]

Molecular weight of [tex]\rm C_6H_1_2O_6[/tex] = 180.156 grams

Mass of water = 125.9 g = 0.1259 kg

Molality = [tex]\rm \frac{45}{180.159}\;\times\;\frac{1}{0.1259}[/tex]

Molality = 1.985 m

The molality of a d-mannose solution prepared by dissolving 45 g of d-mannose in 125.9 g of water has the molality of 1.985 m.

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