Question

Calculate the molality of a solution that is prepared by mixing 25.5 ml of ch3oh (d = 0.792 g/ml and 387 ml of ch3ch2ch2oh (d = 0.811 g/ml.

Answer 1

The solvent is ch3ch2ch2oh with a volume of 387 mL which is equivalent to
387 mL (0.811 g/mL) (1 kg /1000 kg) = 0.3139 kg
The moles of the alchol is
25.5 mL (0.792 g/mL) (1 mol/32 g) = 0.631 mol
The molality is
0.631 mol / 0.3139 kg = 2.01 mol/kg

Answer 2

The molality of solution prepared by mixing 25.5 mL of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ and 387 mL of ${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{OH}}$ is $\boxed{2.01{\text{ m}}}$.

Further Explanation:

Different concentration terms are utilized in determining the concentration of various solutions. Some of the most commonly used terms are written below.

1. Molarity (M)

2. Mole fraction (X)

3. Molality (m)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molality is one of the concentration terms used very often in solutions. It is defined as moles of solute divided by the mass of solvent in kilograms. The formula to calculate molality of solution is as follows:

${\text{Molality of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}$                                     …… (1)

The formula to calculate the density of component is as follows:

${\text{Density of component}} = \dfrac{{{\text{Mass of component}}}}{{{\text{Volume of component}}}}$                                   …… (2)

Rearrange equation (2) to calculate mass of component.

${\text{Mass of component}} = \left( {{\text{Density of component}}} \right)\left( {{\text{Volume of component}}} \right)$            …… (3)

Substitute 25.5 mL for volume of component and 0.792 g/mL for density of component in equation (3) to calculate mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$.

$\begin{aligned} {\text{Mass of C}}{{\text{H}}_{\text{3}}}{\text{OH}} &= \left( {{\text{0}}{\text{.792 g/mL}}} \right)\left( {{\text{25}}{\text{.5 mL}}} \right) \\ & = 20.196{\text{ g}} \\ \end{aligned}$  

Substitute 387 mL for volume of component and 0.811 g/mL for density of component in equation (3) to calculate mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{OH}}$.

$\begin{aligned} {\text{Mass of C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{OH}} &= \left( {{\text{0}}{\text{.811 g/mL}}} \right)\left( {{\text{387 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ kg}}}}{{1{\text{ g}}}}} \right) \\ & = 0.3139{\text{ kg}} \\ \end{aligned}$  

The formula to calculate moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ is as follows:

${\text{Moles of C}}{{\text{H}}_{\text{3}}}{\text{OH}} = \dfrac{{{\text{Mass of C}}{{\text{H}}_{\text{3}}}{\text{OH}}}}{{{\text{Molar mass of C}}{{\text{H}}_{\text{3}}}{\text{OH}}}}$                                                …… (4)

Substitute 20.196 g for mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$  and 32.04 g/mol for molar mass of   in equation (4).

$\begin{aligned} {\text{Moles of C}}{{\text{H}}_{\text{3}}}{\text{OH}} &= \frac{{{\text{20}}{\text{.196 g}}}}{{{\text{32}}{\text{.04 g/mol}}}} \\ &= 0.6303{\text{ mol}} \\ \end{aligned}$  

Substitute 0.6303 mol for moles of solute and 0.3139 kg for mass of solvent in equation (1) to calculate molality of given solution.

$\begin{aligned} {\text{Molality of given solution}} &= \frac{{{\text{0}}{\text{.6303 mol}}}}{{{\text{0}}{\text{.3139 kg}}}} \\ &= 2.007{\text{ m}} \\ & \approx {\text{2}}{\text{.01 m}} \\ \end{aligned}$  

Learn more:

1. Calculation of volume of gas: brainly.com/question/3636135
2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, 2.01 m, molality, moles, mass, 0.3139 kg, CH3OH, CH3CH2CH2OH, 0.6303 mol.