Nalorphene ($C_{19}H_{21}NO_3$), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of $1.5 \times 10^{-3}$ m aqueous solution required for the above dose.

Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of

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C_{19}H_{21}NO_3

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Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

Imagine you're making a cup of sweet tea. You add sugar to hot water, stir, and taste. If you let the tea cool to room temperature, the amount of sugar hasn't changed — but the volume of the liquid has shrunk slightly. If you measured concentration as "grams of sugar per litre of solution," that number would change just because the temperature changed. That's annoying if you're a chemist who needs a reliable, temperature-independent way to describe how much solute is present.

Molality was invented to solve exactly this problem.

The Intuition

Instead of measuring the volume of the solution (which expands and contracts with temperature), molality measures the mass of the solvent. Mass doesn't change with temperature. So molality gives you a concentration that stays the same whether your solution is hot or cold.

Think of it this way:

Molarity = moles of solute per litre of solution (temperature-sensitive)
Molality = moles of solute per kilogram of solvent (temperature-independent)

The solvent is the substance doing the dissolving — usually water. The solute is what gets dissolved — sugar, salt, etc.

The Precise Definition

\text{Molality } (m) = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

The symbol for molality is a lowercase m (not to be confused with M for molarity).

Key points to remember:

The denominator is solvent mass, not solution mass
The unit is mol/kg (often written as simply "m")
It is independent of temperature because mass doesn't change with temperature

Worked Example

Problem: 36 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.

Step 1: Find moles of solute

\text{Moles of glucose} = \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol}

Step 2: Convert solvent mass to kilograms

500 \text{ g} = 0.5 \text{ kg}

Step 3: Apply the formula

m = \frac{0.2 \text{ mol}}{0.5 \text{ kg}} = 0.4 \text{ m}

Note

The answer is 0.4 m (or 0.4 mol/kg). Notice we used the mass of water (500 g), not the mass of the solution (which would be 536 g).

Common Mistake to Avoid

Watch out

Do not use the mass of the solution in the denominator. The formula specifically asks for the mass of the solvent alone. If the problem gives you the total mass of the solution, subtract the mass of the solute to find the solvent mass.

When Do You Use Molality?

Molality is the star in two important situations: …

Why this formula?

Molality Calculation: Why the Formula Works

Molality is a measure of concentration that is temperature-independent — this is its key advantage over molarity. Let's understand why the formula takes the form it does.

The Definition First

Molality ( $m$ ) is defined as:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

The unit is mol/kg, often written as m (e.g., 0.5 m glucose solution).

Why Mass of Solvent, Not Solution?

This is the critical conceptual point.

The Reasoning

Molarity uses volume of solution → volume changes with temperature (expansion/contraction). So molarity changes with temperature.
Molality uses mass of solvent → mass is invariant with temperature. So molality remains constant regardless of temperature changes.

Key insight: By using the solvent's mass (not the solution's volume), we eliminate temperature dependence. This is why molality is preferred for colligative properties (boiling point elevation, freezing point depression) — these properties depend on the number of solute particles, not on temperature.

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

If you have $w_{\text{solute}}$ grams of solute with molar mass $M_{\text{solute}}$ (g/mol):

\text{moles of solute} = \frac{w_{\text{solute}}}{M_{\text{solute}}}

Step 2: Mass of Solvent in kg

If the solvent mass is $W_{\text{solvent}}$ grams:

\text{mass of solvent in kg} = \frac{W_{\text{solvent}}}{1000}

Step 3: Putting It Together

m = \frac{\frac{w_{\text{solute}}}{M_{\text{solute}}}}{\frac{W_{\text{solvent}}}{1000}}

Simplifying:

m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}

The Final Formula (Exam-Ready)

\boxed{m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}}

Where:

$w_{\text{solute}}$ = mass of solute in grams
$M_{\text{solute}}$ = molar mass of solute in g/mol
$W_{\text{solvent}}$ = mass of solvent in grams

Why the ×1000 Factor? …

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Molality relates moles of solute to kilograms of solvent. Given the dose (1.5 mg nalorphene) and molality ( $1.5 \times 10^{-3}$ m), we find the solvent mass needed, then add the negligible solute mass to get total solution mass ≈ 3.23 g.

Understanding Molality

Molality ( $m$ ) measures concentration as moles of solute per kilogram of solvent (not solution). The definition is:

m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}

This problem gives us the dose of nalorphene and the desired molality, asking for the total solution mass. The key insight: we'll first find how much water (solvent) is needed to achieve that molality with 1.5 mg of drug, then add the drug's mass to get the complete solution.

m = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}

Step-by-Step Solution

1. Calculate the molar mass of nalorphene

The molecular formula is $\text{C}_{19}\text{H}_{21}\text{NO}_3$ .

M = 19(12) + 21(1) + 14 + 3(16) = 228 + 21 + 14 + 48 = 311 \text{ g/mol}

2. Find moles of nalorphene in the 1.5 mg dose

Convert the dose to grams: $1.5 \text{ mg} = 1.5 \times 10^{-3} \text{ g}$ .

n = \frac{1.5 \times 10^{-3}}{311} = 4.823 \times 10^{-6} \text{ mol}

3. Use the molality to find the required mass of solvent

Rearrange the molality equation to solve for solvent mass:

m_{\text{solvent (kg)}} = \frac{n_{\text{solute}}}{m}

$m_{\text{solvent (kg)}} = \frac{4.823 \times 10^{-6}}{1.5 \times 10^{-3}} = 3.215 \times 10^{-3} \text{ kg}$ …

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Here are the common mistakes students make when solving this molality-based problem, along with how to avoid each.

1. Confusing Molality ( $m$ ) with Molarity ( $M$ )

The Mistake:

Students treat the given $1.5 \times 10^{-3} \, m$ as molarity and try to use volume (litres) instead of mass of solvent (kg).

Why it’s wrong:

Molality is moles of solute per kg of solvent, not per litre of solution.

How to Avoid:

Always check the unit:

$m$ = mol/kg → molality
$M$ = mol/L → molarity

Write at the top:

Given: $m = 1.5 \times 10^{-3} \, \text{mol/kg}$ (molality)

2. Forgetting to Convert Dose from mg to g

The Mistake:

Using $1.5 \, \text{mg}$ directly in mole calculations without converting to grams.

Why it’s wrong:

Molar mass is in g/mol, so mass must be in grams.

How to Avoid:

Always convert:

1.5 \, \text{mg} = 1.5 \times 10^{-3} \, \text{g}

3. Incorrect Molar Mass Calculation

The Mistake:

Adding atomic masses incorrectly — e.g., forgetting to multiply by the number of atoms.

Why it’s wrong:

$C_{19}H_{21}NO_3$ has 19 carbons, 21 hydrogens, 1 nitrogen, 3 oxygens.

How to Avoid:

Calculate step-by-step:

$C: 19 \times 12 = 228$
$H: 21 \times 1 = 21$
$N: 1 \times 14 = 14$
$O: 3 \times 16 = 48$

Total:

M = 228 + 21 + 14 + 48 = 311 \, \text{g/mol}

4. Using the Wrong Formula for Mass of Solution

The Mistake:

Thinking mass of solution = mass of solvent only, or using $M = n/V$ .

Why it’s wrong:

Solution mass = solute mass + solvent mass.

Molality gives solvent mass, not solution mass.

How to Avoid:

Use the correct sequence:

Find moles of solute:

n = \frac{\text{mass of solute (g)}}{\text{molar mass}} = \frac{1.5 \times 10^{-3}}{311}

n \approx 4.82 \times 10^{-6} \, \text{mol}

Find mass of solvent (in kg) from molality:

m = \frac{n}{\text{mass of solvent (kg)}} \Rightarrow \text{mass of solvent} = \frac{n}{m}

\text{mass of solvent} = \frac{4.82 \times 10^{-6}}{1.5 \times 10^{-3}} \approx 3.21 \times 10^{-3} \, \text{kg} = 3.21 \, \text{g}

Mass of solution = mass of solute + mass of solvent:

= 0.0015 \, \text{g} + 3.21 \, \text{g} \approx 3.2115 \, \text{g}

5. Rounding Too Early

The Mistake:

Rounding intermediate values (like $n$ or solvent mass) before the final step. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GUJCET 2024Set 131 markMCQ
$C_6H_{12}O_6$
›Reveal solutionSolution

Molality is moles solute per kg solvent; set up the equation with (solution − solute) as solvent mass and solve for the solute mass.

Concept: Molar mass of glucose $C_6H_{12}O_6 = 6(12)+12(1)+6(16)=180$ g/mol. Molality $m=\dfrac{\text{moles solute}}{\text{kg solvent}}$ .

Let mass of glucose $=w$ g. Solvent mass $=(2500-w)$ g $=\dfrac{2500-w}{1000}$ kg.
$0.25=\frac{w/180}{(2500-w)/1000}$
$0.25\times\frac{2500-w}{1000}=\frac{w}{180}$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. What is the molality of a 10% w/w aqueous solution of NaOH? (Molecular mass of NaOH = 40 g mol-1)$
(a) 2.78 m
(b) 2.87 m
(c) 2.5 m
(d) 2.05 m
›Reveal solutionSolution

Molality = moles of solute / mass of solvent in kg; for 10% w/w NaOH, take 100 g solution = 10 g NaOH + 90 g water.

Moles of NaOH = 10 g / 40 g mol-1 = 0.25 mol. …

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GUJCET 2021Set 151 markMCQ
$Q. 3.0 gram ethanoic acid in 50 gram benzene having ___ molality? (Atomic weights : H = 1, C = 12, O = 16). (A) 0.1 (B) 1.0 (C) 0.6 (D) 0.06$
›Reveal solutionSolution

$m = \dfrac{\text{mol solute}}{\text{kg solvent}} = \dfrac{0.05}{0.050} = 1.0$ .

Concept: Molality = moles of solute per kg of solvent.

Moles of $CH_3COOH$ (M = 60): $\dfrac{3.0}{60} = 0.05$ mol. …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Molality of 30% w/w aqueous solution of NaOH is -$
(a) 7.5 m
(b) 8.32 m
(c) 10.71 m
(d) 9.17 m
›Reveal solutionSolution

Taking a 100 g basis for a 30% w/w solution gives 30 g NaOH in 70 g water; converting to moles and dividing by the mass of water in kg gives the molality.

Basis: 100 g of solution contains 30 g NaOH and (100-30) = 70 g water = 0.070 kg. …

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GUJCET 2019Set 131 markMCQ
$Q. The value of which of the following unit of concentration will not change with the change in temperature? (A) Formality (B) Normality (C) Molality (D) Molarity$
›Reveal solutionSolution

Molality uses mass, not volume, so it is independent of temperature.

Concept: Concentration units defined using volume (molarity, normality, formality expressed per litre) change with temperature because volume expands or contracts. Molality is defined per kilogram of solvent (mass), and mass does not vary with temperature, so molality is temperature-independent.

Steps: …
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GUJCET 2014Set A1 markMCQ
$Q. What will be the value of molality for an aqueous solution of 10% w/w NaOH. (Na = 23, O = 16, H = 1) (A) 2.778 (B) 5 (C) 10 (D) 2.5$
›Reveal solutionSolution
[!TLDR]

Molality $= 2.778\ \text{mol kg}^{-1}$ .

Concept

Molality $m = \dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}}$ . For a w/w percentage, the stated mass is per 100 g of solution, so the solvent mass is $100 - (\text{solute mass})$ .

Solution

$M(\text{NaOH}) = 23 + 16 + 1 = 40\ \text{g mol}^{-1}$ . …
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GUJCET 2014Set A1 markMCQ
$Q. If 10 ml of 0.1 M aqueous solution of NaCl is divided in to 1000 drops of equal volume, what will be the concentration of one drop? (A) 0.01 M (B) 0.10 M (C) 0.001 M (D) 0.0001 M$
›Reveal solutionSolution
[!TLDR] Concentration is an intensive property; dividing a solution into drops does not change its molarity, so each drop is 0.10 M.

Concept

Molarity $= \dfrac{\text{moles of solute}}{\text{volume of solution (L)}}$ . Both the moles of solute and the volume scale down together when you take a small portion, so their ratio — the concentration — stays the same. Concentration does not depend on how much of the solution you take.

Solution

Total moles $= 0.1\ \text{M} \times 10\times10^{-3}\ \text{L} = 1\times10^{-3}$ mol in 10 mL. …
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←Previous question1.28 — Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃…→Next question1.30 — Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing…

Q.Nalorphene ( $C_{19}H_{21}NO_3$ ), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of $1.5 \times 10^{-3}$ m aqueous solution required for the above dose.

Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

The Intuition

The Precise Definition

Worked Example

Common Mistake to Avoid

When Do You Use Molality?

Why this formula?

Molality Calculation: Why the Formula Works

The Definition First

Why Mass of Solvent, Not Solution?

The Reasoning

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

Step 2: Mass of Solvent in kg

Step 3: Putting It Together

The Final Formula (Exam-Ready)

Why the ×1000 Factor? …

Understanding Molality

Step-by-Step Solution

Method: Molality-to-Mass Conversion using Solute Mass

Steps

1. Confusing Molality ( $m$ ) with Molarity ( $M$ )

2. Forgetting to Convert Dose from mg to g

3. Incorrect Molar Mass Calculation

4. Using the Wrong Formula for Mass of Solution

5. Rounding Too Early

Concept

Solution

Concept

Solution

Q.Nalorphene (C19H21NO3C_{19}H_{21}NO_3C19​H21​NO3​), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5×10−31.5 \times 10^{-3}1.5×10−3 m aqueous solution required for the above dose.

Understanding Molality

Step-by-Step Solution

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Q.Nalorphene ( $C_{19}H_{21}NO_3$ ), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of $1.5 \times 10^{-3}$ m aqueous solution required for the above dose.

Unlock everything free for 14 days