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Some Basic Concepts of Chemistry question

2021 · 27 Jul · Shift 1 · Q14

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Some Basic Concepts of Chemistry
2021 · 27 Jul · Shift 1 · Q14

Some Basic Concepts of Chemistry question

2021 · 27 Jul · Shift 1 · Q14

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1

The density of

NaOH

solution is 1.2 g cm

-

3. The molality of this solution is

\underline{\hspace{2cm}}

m. (Round off to the Nearest Integer) [Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of

H_2O

: 1.0 g cm

-

3]

Numerical answer

View written solutionFree

Correct answer: 5

Interpret the given data

We need the molality of the $\text{NaOH}$ solution.

Given:

Density of solution $= 1.2\ \text{g cm}^{-3}$
Density of water $= 1.0\ \text{g cm}^{-3}$

Since no concentration is directly given, we use the density increase to infer dissolved solute mass in a convenient volume.

Take $1\ \text{cm}^3$ of solution

Mass of $1\ \text{cm}^3$ solution: $1.2\ \text{g}$

If the same volume were pure water, its mass would be: $1.0\ \text{g}$

So, mass of $\text{NaOH}$ dissolved in $1\ \text{cm}^3$ solution is taken as: $1.2 - 1.0 = 0.2\ \text{g}$

Hence, in $1\ \text{cm}^3$ solution:

mass of solute $= 0.2\ \text{g}$
mass of solvent (water) $= 1.0\ \text{g}$

Find moles of $\text{NaOH}$

Molar mass of $\text{NaOH}$ : $23 + 16 + 1 = 40\ \text{g mol}^{-1}$

Moles of $\text{NaOH}$ : $n = \frac{0.2}{40} = 0.005\ \text{mol}$

Convert mass of solvent into kg

Mass of water: $1.0\ \text{g} = 0.001\ \text{kg}$

Calculate molality

Molality is: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

So, $m = \frac{0.005}{0.001} = 5$

Final answer

The molality of the solution is: $\boxed{5\ \text{m}}$

Rounded to the nearest integer: $\boxed{5}$

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