Question

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? Assume density of water is 1.00 g/ml.

Answer 1

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

Answer 2

The molality of fructose in the solution is  0.531 molal.

What is molality?

Molality is the measure of the moles of any solute in a solution per unit kg of the solvent.

Given the mass of fructose is 3.35g

Volume is 35.0 ml

Density of water is 1.00 g/ml

Calculate the number of moles of fructose

$\rm Number\;of \;moles= \dfrac{mass}{molar\;mass}\\\\\rm Number\;of \;moles= \dfrac{ 3.35g}{180.16 g/mol} = 0.0186\; mol$

Now, calculate the molality

First, convert the volume of water into kg

35.0 ml × ( 1 g/mL )  × ( Kg/1000g) = 0.035 Kg H₂O

$\rm Molality = \dfrac{mol}{kg}\\\\\rm Molality = \dfrac{0.0186\;mol}{0.035 kg}= 0.531\; molal$

Thus, the molality of fructose is 0.531 molal.

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