"Molality $(\mathrm{m})$ of $3 \mathrm{M}$ aqueous solution of $\mathrm{NaCl}$ is : (Given : Density of solution $=1.25 \mathrm{~g} \mathrm{~mL}^{-1}$, Molar mass in $\mathrm{g} \mathrm{~mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5$)"

Question

Accepted Answer

"

To calculate the molality $(\mathrm{m})$ of a 3M (molar) aqueous solution of $\mathrm{NaCl}$, we need to follow certain steps, using the given data and understanding the definitions properly.

Molarity (M) is defined as the number of moles of solute per liter of solution. Molality $(\mathrm{m})$, on the other hand, is defined as the number of moles of solute per kilogram of solvent. The given density of the solution is $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$ which allows us to calculate the mass of the solvent in a given volume of solution.

We know:

Molarity (M) of NaCl = $3 \mathrm{~M}$
Density of NaCl solution = $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$
Molar mass of $\mathrm{NaCl} = \mathrm{Na} + \mathrm{Cl} = 23 + 35.5 = 58.5 \mathrm{~g} \mathrm{~mol}^{-1}$

First, to find the molality, we need to find the moles of NaCl in a given mass of the solvent. Keeping in mind that $3 \mathrm{~M}$ means there are $3 \mathrm{~moles}$ of $\mathrm{NaCl}$ per liter of solution, we consider 1 liter (or 1000 mL) of solution to simplify our calculation, knowing that density can convert volume to mass directly.

The mass of 1 liter of NaCl solution is $1.25 \mathrm{~g} \mathrm{~mL}^{-1} \times 1000 \mathrm{~mL} = 1250 \mathrm{~g}$.

Since we have a 3 M solution of NaCl,

$3 \mathrm{~moles} = \frac{\text{mass of the solution} - \text{mass of solvent}}{\text{molar mass of NaCl}}$

With 3 moles of NaCl, the mass of NaCl present in 1 L solution $= 3 \mathrm{~moles} \times 58.5 \mathrm{~g/mol} = 175.5 \mathrm{~g}$

The mass of the solvent (water) can be found by subtracting the mass of NaCl from the total mass of the solution.

$\text{Mass of solvent} = 1250 \mathrm{~g} - 175.5 \mathrm{~g} = 1074.5 \mathrm{~g}$

To convert this mass into kilograms (since molality is expressed per kilogram of solvent),

$1074.5 \mathrm{~g} = 1.0745 \mathrm{~kg}$

Finally, the molality $(\mathrm{m})$,

$\mathrm{m} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{3 \mathrm{~moles}}{1.0745 \mathrm{~kg}}$

$\mathrm{m} \approx 2.79$

So, the correct answer is

Option D: 2.79 m

"

Molality $(\mathrm{m})$ of $3 \mathrm{M}$ aqueous solution of $\mathrm{NaCl}$ is : (Given : Density of solution $=1.25 \mathrm{~g} \mathrm{~mL}^{-1}$, Molar mass in $\mathrm{g} \mathrm{~mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5$)

Solution

About this question

Drill 25 more like these. Every day. Free.

Molality $(\mathrm{m})$ of $3 \mathrm{M}$ aqueous solution of $\mathrm{NaCl}$ is : (Given : Density of solution $=1.25 \mathrm{~g} \mathrm{~mL}^{-1}$, Molar mass in $\mathrm{g} \mathrm{~mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5$)

Solution

About this question

More Some Basic Concepts of Chemistry questions

Drill 25 more like these. Every day. Free.