Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is
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Q: Why can't we just assume the mass of the solvent is 1 kg?

Because molarity is defined per liter of the *entire solution*, not per kilogram of the solvent. The solution contains both the solvent and the solute, so 1 Liter of solution does not equal 1 kg of solvent.

Q: How did we get 60 g/mol for the molar mass of acetic acid?

The chemical formula for acetic acid is $CH_3COOH$. Adding the atomic masses: $2 \times 12$ (Carbon) $+ 4 \times 1$ (Hydrogen) $+ 2 \times 16$ (Oxygen) $= 24 + 4 + 32 = 60\text{ g/mol}$.

Q: Why do we subtract $M \times M_1$ in the denominator of the formula?

The term $1000d$ gives the mass of $1000\text{ mL}$ of the solution. The term $M \times M_1$ gives the mass of the solute in that $1000\text{ mL}$. Subtracting them gives the mass of the pure solvent in grams, which is required for molality.

Q: Is molality dependent on temperature?

No! Molality is defined using mass (moles of solute per kg of solvent). Since mass does not change with temperature, molality remains constant. Molarity, however, uses volume, which expands or contracts with temperature.

Q: Can I solve this without memorizing the formula?

Absolutely! Just assume you have $1\text{ L}$ of solution. Find its mass using density, find the mass of the solute using moles and molar mass, subtract to find the mass of the solvent, and divide the moles by the solvent mass in kg.

Question

Density of a $2.05	ext{ M}$ solution of acetic acid in water is $1.02	ext{ g /mL}$. The molality of the solution is

Options:

(A) $1.14	ext{ mol kg}^{-1}$

(B) $3.28	ext{ mol kg}^{-1}$

(C) $2.28	ext{ mol kg}^{-1}$

(D) $0.44	ext{ mol kg}^{-1}$

Sigma Percentile · Accepted Answer

### Decoding the Concentration Puzzle

Imagine you are standing in a chemistry lab, holding a beaker filled with an aqueous solution of acetic acid ($CH_3COOH$). You are given two pieces of information: the molarity ($M$) is $2.05	ext{ M}$, and the density ($d$) is $1.02	ext{ g/mL}$. Your mission is to find the molality ($m$) of this solution.

At first glance, molarity and molality sound similar, but they measure fundamentally different things. Molarity tells us the number of moles of solute per **liter of solution**, while molality tells us the number of moles of solute per **kilogram of solvent**. To bridge the gap between volume (solution) and mass (solvent), we need our trusty bridge: density.

### The Intuitive Approach (No Formula Required!)

Let's break this down logically without memorizing any formulas.

Assume we have exactly $1	ext{ Liter}$ ($1000	ext{ mL}$) of this solution. 
Because the molarity is $2.05	ext{ M}$, this $1	ext{ L}$ solution contains exactly $2.05	ext{ moles}$ of acetic acid.

What is the mass of this $1	ext{ L}$ solution? 
Using the density, $	ext{Mass} = 	ext{Volume} 	imes 	ext{Density} = 1000	ext{ mL} 	imes 1.02	ext{ g/mL} = 1020	ext{ g}$.

Now, what is the mass of the solute inside it? 
The molar mass of acetic acid ($CH_3COOH$) is $60	ext{ g/mol}$. 
So, $	ext{Mass of solute} = 2.05	ext{ moles} 	imes 60	ext{ g/mol} = 123	ext{ g}$.

To find molality, we need the mass of the **solvent** (water). 
$	ext{Mass of solvent} = 	ext{Total Mass of Solution} - 	ext{Mass of Solute}$
$	ext{Mass of solvent} = 1020	ext{ g} - 123	ext{ g} = 897	ext{ g} = 0.897	ext{ kg}$.

Finally, molality is moles of solute divided by mass of solvent in kg:
$$m = \frac{2.05	ext{ moles}}{0.897	ext{ kg}} \approx 2.285	ext{ mol kg}^{-1}$$

### The Master Formula Shortcut

While the intuitive method is beautiful, competitive exams demand speed. We can condense the entire logical process above into one powerful master equation:

$$m = \frac{M 	imes 1000}{1000d - M \cdot M_1}$$

Here, $M$ is molarity, $d$ is density, and $M_1$ is the molar mass of the solute. Notice how the denominator $(1000d - M \cdot M_1)$ is exactly what we did to find the mass of the solvent in grams!

### Final Calculation

Let's plug in our values:
$$m = \frac{2.05 	imes 1000}{1000(1.02) - 2.05(60)}$$
$$m = \frac{2050}{1020 - 123}$$
$$m = \frac{2050}{897} \approx 2.28	ext{ mol kg}^{-1}$$

**Concept Check:** Why do we care about molality? Because unlike molarity, which changes as liquids expand or contract with temperature, molality is based purely on mass. It is completely **temperature independent**, making it the preferred concentration term for colligative properties like boiling point elevation and freezing point depression!

Animated Solution for Chemistry - Basic Concepts in Chemistry: Density of a $2.05 M$ solution of acetic acid in water is $1.02 g /mL$ . The molality of the solution is

Select Answer:

Visualized Solution

\text{Given Data}

\text{Relation between } m \text{ and } M

\text{Molar Mass of } CH_3COOH

\text{Substitution}

\text{Simplification}

\text{Final Calculation}

\text{Concept Check}

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Decoding the Concentration Puzzle

The Intuitive Approach (No Formula Required!)

The Master Formula Shortcut

Final Calculation

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Animated Solution for Chemistry - Basic Concepts in Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Density of a 2.05 M solution of acetic acid in water is 1.02 g /mL. The molality of the solution is