Question

A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ), molality ( m ), and mass percent concentration of the solution

Answer 1

Answer:

[KOH] = 1.47 M

[KOH] = 1.22 m

KOH = 6.86 % m/m

Explanation:

Let's analyse the data

1.87 L is the volume of solution

Density is 1.29 g/mL → Solution density

155 g of KOH → Mass of solute

Moles of solute is (mass / molar mass) = 2.76 moles.

Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M

Let's determine, the mass of solvent.

Molality is mol of solute / 1kg of solvent

We can use density to find out the mass of solution

Mass of solution - Mass of solute = Mass of solvent

Density = Mass / volume

1.29 g/mL = Mass / 1870 mL

Notice, we had to convert L to mL, cause the units of density.

1.29 g/mL . 1870 mL = Mass → 2412.3 g

2412.3 g - 155 g = 2257.3 g of solvent

Let's convert the mass of solvent to kg

2257.3 g / 1000 = 2.25kg

2.76 mol / 2.25kg = 1.22 m (molality)

% percent by mass = mass of solute in 100g of solution.

(155 g / 2257.3 g) . 100g = 6.86 % m/m

Answer 2

The molar concentration of a solution is the estimation of the chemical component in the selective solute of a solution while molality is the number of moles of a solute in a given solution.

Given,

• The volume of the solution = 1.87 L

• Mass of the KOH = 155 g

• Density of the solution = 1.29 g/mL

• Molar mass of KOH = 56.10 g/mol

To calculate moles of the solute the following formula is used:

$\text {Moles of solute} &= \dfrac{\text{mass}}{\text{molar mass}}$

$\text {Moles of solute} &= \dfrac{155 \text{g} }{56.10 \text{g/mol}} & = 2.76 \; \text{moles}$

1. To calculate Molarity (M) the following formula is used:

$\text{Molarity} & = \dfrac{\text{n}}{\text{V}}$

where n is the number of moles, and v is the volume of solution in Litres.

$\text{Molarity} & = \dfrac{\text{2.76 mol}}{\text{1.87 L }} & = 1.47 \; \text{M}$

2. To calculate Molality (m) the following formula is used:

$\text{Molality(m)} & = \dfrac{\text{moles of solute}}{\text{mass of solvent in kgs}}$

To know the mass of solvent:

Mass of solution - Mass of solute = Mass of solvent

$\text{Density} &= \dfrac{\text{ Mass}}{\text {Volume}}$

mass = density × volume

=1.29 g/mL × 1870 mL

= 2412.3 g

2412.3 g ( Mass of solution) - 155 g (Mass of solute)

= 2257.3 g of solvent

Now convert the mass of solvent into kg:

$\dfrac{2257.3 \text{g}}{ 1000 } & = 2.25 \text{kg}$

$\text{Molality(m)} & = \dfrac{\text{ 2.76 mol}}{\text{ 2.25 kg}} &= 1.22 \text{m}$

3. To calculate the percentage by mass:

The mass of the element is divided by the mass of the solute in the compound.

Mass of solute in 100 g of solution can be calculated as:

$\dfrac{155 \; \text{g}}{2257.3 \; \text{g} } \times 100$

= 6.86 % m/m

Therefore, molarity  1.47 M, molality 1.22 m and  mass percent concentration of the solution is 6.86 % m/m.

To learn more about molarity, molality and mass percentage refer to the link:

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