A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ i

Q: Why do we use the mass of the sodium ion and not the whole sodium sulphate molecule?

The question specifically asks for the molality of the $\mathrm{Na}^+$ ions. Therefore, we must treat the sodium ions as our primary solute and use their specific mass and molar mass.

Q: Does the presence of sulphate ions affect the molality of the sodium ions?

No. Molality is a measure of the concentration of a specific solute. The concentration of sodium ions depends only on the moles of sodium ions and the mass of the solvent, regardless of what other ions are floating around.

Q: Why is molality preferred over molarity in some scientific calculations?

Molality is based on mass, which does not change with temperature. Molarity is based on volume, which expands or contracts with temperature changes, making molality a more reliable unit for experiments involving temperature variations.

Q: What is the molar mass of a sodium ion? Is it different from a neutral sodium atom?

An ion is just an atom that has lost or gained an electron. Since the mass of an electron is practically negligible, the molar mass of a $\mathrm{Na}^+$ ion is identical to that of a neutral $\mathrm{Na}$ atom, which is $23\,\mathrm{g/mol}$.

Q: What if the solvent mass was given in liters instead of kilograms?

For pure water, the density is approximately $1\,\mathrm{kg/L}$, so $1\,\mathrm{L}$ of water is roughly $1\,\mathrm{kg}$. However, for other solvents, you must use their specific density to convert the given volume into mass before calculating molality.

Question

A solution of sodium sulphate contains $92\,\mathrm{g}$ of $\mathrm{Na}^+$ ions per kilogram of water. The molality of $\mathrm{Na}^+$ ions in that solution in $\mathrm{mol\,kg}^{-1}$ is

Options:

(A) $16$

(B) $4$

(C) $132$

(D) $8$

Sigma Percentile · Accepted Answer

The concept of concentration is the heartbeat of solution chemistry. Whether you are brewing a perfect cup of coffee or calculating the exact amount of reagent needed for a titration, knowing *how much* stuff is dissolved in *how much* liquid is crucial. In this problem, we are tasked with finding the molality of a specific ion floating in a sea of water.

### Analyzing the Setup

Imagine you are standing in a laboratory. On the bench in front of you is a beaker containing exactly $1\,\mathrm{kg}$ of pure water. Into this water, someone has dissolved a quantity of sodium sulphate ($\mathrm{Na}_2\mathrm{SO}_4$).

We don't know the exact mass of the sodium sulphate added, but we are given a very specific piece of information: the solution contains exactly $92\,\mathrm{g}$ of sodium ions ($\mathrm{Na}^+$). Our mission is to determine the molality of these sodium ions.

### The Master Equation

To solve this, we need to bring out our trusty formula for molality. Molality ($m$) is defined as the number of moles of solute divided by the mass of the solvent in kilograms.

$$m = \frac{n_{\mathrm{solute}}}{W_{\mathrm{solvent}}(\mathrm{kg})}$$

Unlike molarity, which depends on the volume of the solution, molality depends strictly on the mass of the solvent. This makes it incredibly useful because mass does not change with temperature!

### Final Calculation

First, we need to find the number of moles of our solute, the sodium ions. We know that the molar mass of sodium is $23\,\mathrm{g/mol}$.

$$n_{\mathrm{Na}^+} = \frac{	ext{Given Mass}}{	ext{Molar Mass}} = \frac{92\,\mathrm{g}}{23\,\mathrm{g/mol}} = 4\,\mathrm{mol}$$

We have exactly $4\,\mathrm{mol}$ of sodium ions. Now, we substitute this back into our molality equation. The mass of our solvent (water) is already given as $1\,\mathrm{kg}$.

$$m = \frac{4\,\mathrm{mol}}{1\,\mathrm{kg}} = 4\,\mathrm{mol\,kg}^{-1}$$

And there we have it! The molality of the sodium ions is **$4\,\mathrm{mol\,kg}^{-1}$**.

### The "What If" Scenario

Let's push our understanding a bit further. What if the question had asked for the molality of the parent salt, sodium sulphate ($\mathrm{Na}_2\mathrm{SO}_4$), instead of just the sodium ions?

When sodium sulphate dissolves, it dissociates completely:
$$\mathrm{Na}_2\mathrm{SO}_4 ightarrow 2\mathrm{Na}^+ + \mathrm{SO}_4^{2-}$$

Notice that one molecule of sodium sulphate produces **two** sodium ions. Therefore, the number of moles of the parent salt must be exactly half the number of moles of the sodium ions.

$$m_{\mathrm{Na}_2\mathrm{SO}_4} = \frac{m_{\mathrm{Na}^+}}{2} = 2\,\mathrm{mol\,kg}^{-1}$$

Always pay close attention to exactly *what* species the question is asking about. A small detail like this can completely change your final answer!

Animated Solution for Chemistry - Basic Concepts in Chemistry: A solution of sodium sulphate contains $92 g$ of $Na^{+}$ ions per kilogram of water. The molality of $Na^{+}$ ions in that solution in $mol kg^{- 1}$ is

Select Answer:

Visualized Solution

w_{\mathrm{Na}^+} = 92\,\mathrm{g}

m = \frac{n}{W}

n_{\mathrm{Na}^+} = \frac{92}{23}

n_{\mathrm{Na}^+} = 4\,\mathrm{mol}

m = 4\,\mathrm{mol\,kg}^{-1}

m_{\mathrm{Na}_2\mathrm{SO}_4} = ?

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Analyzing the Setup

The Master Equation

Final Calculation

The "What If" Scenario

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Select Answer:

Visualized Solution

w_{\mathrm{Na}^+} = 92\,\mathrm{g}

m = \frac{n}{W}

n_{\mathrm{Na}^+} = \frac{92}{23}

n_{\mathrm{Na}^+} = 4\,\mathrm{mol}

m = 4\,\mathrm{mol\,kg}^{-1}

m_{\mathrm{Na}_2\mathrm{SO}_4} = ?

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Analyzing the Setup

The Master Equation

Final Calculation

The "What If" Scenario

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