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love history [14]
3 years ago
10

A solution is made by dissolving 3.35 g of fructose in 35.0 ml of water. What is the molality of fructose in the solution? Assum

e density of water is 1.00 g/ml.
Chemistry
1 answer:
ludmilkaskok [199]3 years ago
6 0

Answer:

m = 0.531 molal

Explanation:

∴ m fructose = 3.35 g

∴ V water = 35.0 mL

∴ ρ H2O = 1 g/mL

  • molality = moles solute / Kg solvent

∴ Mw fructose = 180.16 g/mol

⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose

⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O

⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O

⇒ m = 0.531 molal

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In 1 L solution the number of sucrose moles are - 0.758 mol
Therefore in 1.55 L solution, sucrose moles are - 0.758 mol/L x 1.55 L
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