Question

41. A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.

Answer 1

Answer:

1.08 m is the molality of the solution.

Explanation:

1st step: Determine the mass of solvent.

13% by mass means that 13 g are contained in 100 g of solution.

Solution mass = Solute mass + solvent mass

100 g = 13 g + Solvent mass

Solvent mass = 100 g - 13 g → 87 g

Second Step: Let's determine the moles of our solute mass (mass / molar mass)

13 g / 138.2 g/mol = 0.094 moles

Third step: Replace the data to determine molality.

Molality are the moles of solute in 1kg of solvent (mol/kg)

Before this, we must convert the solvent mass to kg

87 g . 1kg / 1000 g = 0.087 kg

molality → 0.094 mol / 0.087 kg = 1.08 m

Answer 2

The molality of a 13.0% solution of $\rm K_2CO_3$ solution is 1.08 m.

What is molality?

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

Step 1: Calculate the mass of the solvents

13.0% solution means 13 g of $\rm K_2CO_3$ is in the 100 g solution

Solution mass = Solute mass + solvent mass

100 g = 13 g + Solvent mass

Solvent mass = 100 g - 13 g = 87 g

Step 2: Calculate the moles

$\rm Number\;of \;moles= \dfrac{mass}{molar\;mass}\\\\\\Number\;of \;moles= \dfrac{13\;g}{138.2\;g/mol}= 0.094\;moles$

Step 3: Calculate the molality

Molality is the moles of solute in 1 kg of solvent (mol/kg)

Convert the mass of solvent into grams

87 g 1kg  = 0.087 kg

$\rm Molality = \dfrac{mol}{kg} \\\\\\Molality = \dfrac{0.094\;mol}{0.087\;kg}= 1.08\;m$

Thus, the molality of the solution is 1.08 m.

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