A measured temperature on Fahrenheit scale is 200 °F. What will this reading be on Celsius scale? (i) 40 °C (i

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Q. A measured temperature on Fahrenheit scale is 200 °F. What will this reading be on Celsius scale?

(i) 40 °C

(ii) 94 °C

(iii) 93.3 °C

(iv) 30 °C

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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:

Colligative properties — properties like boiling point elevation and freezing point depression depend on the number of solute particles per mass of solvent, not per volume. Molality is the natural choice here.
Temperature-varying experiments — if you're working at different temperatures, molality keeps your concentration constant while molarity would drift.

Quick Comparison: Molarity vs Molality

Property	Molarity (M)	Molality (m)
Definition	moles solute / L solution	moles solute / kg solvent
Depends on temperature?	Yes (volume changes)	No (mass is constant)
Common unit	mol/L	mol/kg
Best used for	Room-temp reactions, titrations	Colligative properties, temperature studies

Final Takeaway

Molality is the concentration measure that stays honest when temperature changes. It's moles of solute per kilogram of solvent — and that's the whole story. Once you remember that the denominator is solvent mass, not solution volume, you've got it.

"Molality formula and calculation examples" and "molarity vs molality class 12 chemistry" are frequently searched terms, both grounded in the Solutions chapter of the NCERT/CBSE Class 12 Chemistry curriculum. Molality-based numericals are a near-guaranteed question type in board exams and JEE Main colligative-properties problems.

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?

Because molality requires solvent mass in kg, but we usually measure it in grams. The factor 1000 converts grams to kilograms:

1 \text{ kg} = 1000 \text{ g}

So if solvent mass is in grams, we multiply by 1000 to get the correct denominator in kg.

Common Mistake to Avoid

Do not use the mass of the solution (solute + solvent) in the denominator. The formula specifically requires mass of solvent only.

Example: If you dissolve 10 g NaCl in 90 g water, the solvent mass is 90 g, not 100 g.

Quick Check: Why This Matters in Exams

In problems involving:

Freezing point depression: $\Delta T_f = K_f \times m$
Boiling point elevation: $\Delta T_b = K_b \times m$

You must use molality, not molarity. The formula above is how you calculate $m$ from given masses.

Bottom line: Molality = moles of solute per kg of solvent. The ×1000 factor is just a unit conversion. The real conceptual leap is understanding why we use solvent mass — for temperature independence.

Convert Fahrenheit to Celsius using the linear relationship between the two scales; 200 °F = 93.3 °C.

Temperature scales are human constructs that assign numbers to the physical sensation of hot and cold. The Fahrenheit and Celsius scales differ in both their zero points and the size of their degree intervals. Fahrenheit sets water's freezing point at 32 °F and boiling at 212 °F (a 180-degree span), while Celsius uses 0 °C and 100 °C (a 100-degree span). The conversion formula captures this linear relationship.

C = \frac{5}{9}(F - 32)

This formula works because we first shift the Fahrenheit reading down by 32 to align the zero points, then scale by $\frac{5}{9}$ to account for the different degree sizes (since 180 Fahrenheit degrees equal 100 Celsius degrees, and $\frac{100}{180} = \frac{5}{9}$ ).

Step-by-step conversion:

Identify the given temperature.

We have $F = 200$ °F.
Subtract the offset.

The Fahrenheit scale is shifted by 32 degrees relative to Celsius at the freezing point of water:

F - 32 = 200 - 32 = 168

Apply the scaling factor. Since Fahrenheit degrees are smaller than Celsius degrees (it takes 1.8 °F to equal 1 °C), we multiply by $\frac{5}{9}$ :

C = \frac{5}{9} \times 168

Calculate the result.

C = \frac{5 \times 168}{9} = \frac{840}{9} = 93.\overline{3} \text{ °C}

This is exactly $93.3$ °C (with the 3 repeating).

Watch out

A common mistake is to use $\frac{9}{5}$ instead of $\frac{5}{9}$ , or to forget the subtraction of 32. Remember: Fahrenheit → Celsius requires subtracting 32 first, then multiplying by $\frac{5}{9}$ . The reverse (Celsius → Fahrenheit) uses $F = \frac{9}{5}C + 32$ .

✓Final answer

The correct option is (iii) 93.3 °C .

Here are the common mistakes students make when converting 200 °F to Celsius, along with how to avoid each.

Mistake 1: Using the Wrong Formula (Inverting the Relationship)

The Mistake: Students often confuse the conversion formulas. They might use $C = \frac{9}{5}F + 32$ (which is the formula to convert from Celsius to Fahrenheit) instead of the correct one.
Why it happens: Memorizing formulas without understanding the logic of the scale intervals.
How to Avoid:
- Remember the logic: The Celsius scale has 100 degrees between freezing (0°C) and boiling (100°C). The Fahrenheit scale has 180 degrees between freezing (32°F) and boiling (212°F).
- The Ratio: A change of 1°C equals a change of 1.8°F (or $\frac{9}{5}$ °F). Therefore, to go from °F to °C, you must first subtract the offset (32) and then divide by 1.8 (or multiply by $\frac{5}{9}$ ).
- Correct Formula:

C = \frac{5}{9}(F - 32)

- **Quick Check:** If you use the wrong formula ($C = \frac{9}{5}(200) + 32$), you get 392°C, which is absurdly high. This instantly tells you the formula is wrong.

Mistake 2: Forgetting to Subtract 32 First

The Mistake: Students directly multiply the Fahrenheit value by $\frac{5}{9}$ without subtracting 32. For example: $C = \frac{5}{9} \times 200 \approx 111.1^\circ C$ .
Why it happens: Rushing through the steps or treating the formula as a simple multiplication.
How to Avoid:
- Follow the order of operations strictly. The formula is $C = \frac{5}{9}(F - 32)$ . The subtraction inside the bracket is the first step.
- Step-by-step:
  1. Subtract 32: $200 - 32 = 168$
  2. Multiply by $\frac{5}{9}$ : $168 \times \frac{5}{9} = \frac{840}{9} = 93.33...$
- Result: $\mathbf{93.3^\circ C}$ (Option (iii)).

Mistake 3: Incorrect Arithmetic with the Fraction $\frac{5}{9}$

The Mistake: Students make errors when dividing by 9 or multiplying by 5. For instance, they might calculate $168 \div 9 = 18.66$ and then forget to multiply by 5, getting 18.7°C. Or they might incorrectly compute $168 \times 5 = 740$ instead of 840.
Why it happens: Careless calculation or not simplifying the fraction.
How to Avoid:
- Simplify before multiplying: Check if the number (after subtracting 32) is divisible by 9. In this case, $168 \div 9 = 18.666...$ (not a whole number), so you must do the full multiplication.
- Do the multiplication first: $168 \times 5 = 840$ . Then divide: $840 \div 9 = 93.33...$
- Use decimal approximation: $\frac{5}{9} \approx 0.5556$ . So $168 \times 0.5556 \approx 93.34^\circ C$ . This confirms the answer.

Mistake 4: Confusing the Answer with a Nearby Trap Option

The Mistake: Students get an answer like 93.3°C but then see option (ii) 94°C and select it, thinking it's "close enough" or that they rounded incorrectly.
Why it happens: Not trusting the exact calculation or misreading the options.
How to Avoid:
- Calculate precisely: The exact value is $93.\overline{3}^\circ C$ . The option (iii) is 93.3 °C, which is the correct rounded form.
- Recognize trap options: Option (ii) 94°C is a common rounding error (rounding 93.33 up to 94). Option (i) 40°C is what you get if you mistakenly use $C = F - 32$ (200 - 32 = 168, then wildly wrong). Option (iv) 30°C is a random low number.
- Rule of thumb: For a high Fahrenheit value like 200°F, the Celsius equivalent should be high (near boiling point of water, 100°C). 93.3°C makes physical sense.

Summary Table for Quick Revision

Mistake	Wrong Calculation	Correct Step	Final Answer
Wrong Formula	$C = \frac{9}{5}(200) + 32 = 392$	Use $C = \frac{5}{9}(F - 32)$	93.3°C
Forgot to Subtract 32	$C = \frac{5}{9}(200) = 111.1$	First: $200 - 32 = 168$	93.3°C
Arithmetic Error	$168 \times 5 = 740$	$168 \times 5 = 840$	93.3°C
Picked Trap Option	93.33 → rounded to 94	Exact value is $93.\overline{3}$	93.3 °C (Option (iii))

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

Showing the 12 most recent of 31 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$2.0$
›Reveal solutionSolution
A 2.0 molal solution means 2 moles of solute in 1 kg of water. Convert the solvent mass to moles, then apply the mole fraction formula: $\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$ . The answer is $\boxed{0.0347}$ .

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. This distinction matters because we need to count the moles of both solute and solvent separately to find the mole fraction.

When we say a solution is 2.0 molal, we're saying there are 2.0 moles of solute dissolved in exactly 1000 g (1 kg) of water. The mole fraction then asks: what fraction of the total number of particles (molecules) in the solution comes from the solute?

Let me work through the calculation systematically.

1. Identify what we know from "2.0 molal aqueous solution"

The molality $m = 2.0$ tells us:

Moles of solute: $n_{\text{solute}} = 2.0 \text{ mol}$

Mass of water (solvent): $1000 \text{ g}$

2. Convert the mass of water to moles

Water has a molar mass of $18 \text{ g/mol}$ , so:
$n_{\text{water}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = 55.56 \text{ mol}$
3. Calculate the total moles in the solution
$n_{\text{total}} = n_{\text{solute}} + n_{\text{water}} = 2.0 + 55.56 = 57.56 \text{ mol}$
4. Apply the mole fraction formula

The mole fraction of solute is:
$\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{total}}} = \frac{2.0}{57.56} = 0.03474$
Rounding to three significant figures: $\chi_{\text{solute}} = 0.0347$

Watch out

A common mistake is to confuse molality (moles per kg of solvent) with molarity (moles per liter of solution). For molality, the denominator mass refers only to the solvent, which is why we can directly count 1000 g of water here.

Tip

For dilute aqueous solutions, you can use the quick approximation $\chi_{\text{solute}} \approx \frac{m}{55.56}$ where $m$ is the molality, since water contributes roughly 55.56 moles per kg. Here: $\frac{2.0}{55.56} \approx 0.036$ , close to our answer.

✓Final answer
$0.0347$
CBSE 2026Set A1 markMCQ
$Q. 34.2 g of sugar is present in 234.2 g of its aqueous solution. Then its molal concentration is$
(a) 0.1
(b) 0.5
(c) 5.5
(d) 55.0
›Reveal solutionSolution

Moles of sugar = 34.2/342 = 0.1; mass of solvent (water) = 234.2 - 34.2 = 200 g = 0.2 kg; molality = 0.1/0.2 = 0.5 m.

Molar mass of sugar (sucrose) = 342 g/mol.

Moles of sugar = 34.2/342 = 0.1 mol.

Mass of solution = 234.2 g, mass of sugar = 34.2 g, so mass of water = 234.2 - 34.2 = 200 g = 0.2 kg.

Molality = moles of solute / mass of solvent in kg = 0.1/0.2 = 0.5 mol/kg.

✓Final answer
$(b) 0.5.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. What will be the molarity of 30 ml of 0.5 M H2SO4 solution diluted to 50 ml?$
(a) 0.3 M
(b) 0.03 M
(c) 3 M
(d) 0.13 M
›Reveal solutionSolution

Dilution does not change the number of moles of solute, so M1V1 (before) = M2V2 (after).

Given: M1 = 0.5 M, V1 = 30 mL, final volume V2 = 50 mL.

Using the dilution law: M1V1 = M2V2

0.5 x 30 = M2 x 50

15 = 50 x M2

M2 = 15/50 = 0.3 M

✓Final answer
$(a) 0.3 M.$
CBSE 2026Set ANNUAL1 mark
$Q.Define mole fraction.$
›Reveal solutionSolution

Mole fraction expresses a component's amount relative to the total moles in a mixture, independent of temperature.

For a solution containing components with n1, n2, n3, ... moles, the mole fraction of component 1 is defined as:

x1 = n1 / (n1 + n2 + n3 + ...)

It is a dimensionless quantity, and the mole fractions of all components in a solution always sum to 1. Unlike molarity, mole fraction does not depend on temperature, since it is based purely on the number of moles, not volume.

✓Final answer
$Mole fraction = (moles of a given component) / (total moles of all components in the solution).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. In an acid-base titrimetric analysis, the concentration of a sulphuric acid analyte is found to be 0.044 M. The strength of the acid in g/L is -$
(a) 0.44
(b) 4.31
(c) 2.15
(d) 44.00
›Reveal solutionSolution

Strength in g/L = molarity × molar mass = 0.044 × 98 ≈ 4.31 g/L, so option (B).

The strength of a solution in grams per litre is related to its molarity by

$\text{Strength (g/L)} = \text{Molarity (mol/L)} \times \text{Molar mass (g/mol)}.$

The molar mass of sulphuric acid $H_2SO_4 = 2(1) + 32 + 4(16) = 98\ g\,mol^{-1}$ .

$\text{Strength} = 0.044 \times 98 = 4.312 \approx 4.31\ g\,L^{-1}.$

✓Final answer
$(B) 4.31 g/L.$
CBSE 2025Set ANNUAL1 mark
$Q. Write the definition of molality.$
›Reveal solutionSolution

Molality expresses concentration as moles of solute per kilogram of SOLVENT (not solution), and unlike molarity it does not change with temperature.

Definition:

Molality (m) = (moles of solute) / (mass of solvent in kg)

Unit: mol kg^-1 (also written 'molal', symbol m)

Because molality is defined using the MASS of solvent (which does not change with temperature) rather than the VOLUME of solution (which expands/contracts with temperature), molality is a temperature-independent measure of concentration - unlike molarity, which does vary with temperature.

✓Final answer
$Molality = moles of solute per kilogram of solvent.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the molarity of a solution with a mass of solute 10 kg mass and 100 litre volume?$
(a) 0.1 molar
(b) 1 molar
(c) 10 molar
(d) 100 molar
›Reveal solutionSolution

Molarity is defined as moles of solute per litre of solution; dividing the given amount of solute by the solution volume gives 0.1 M.

Molarity is defined as:

$M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$

Note: as printed, the question states the solute quantity as '10 kg mass'; for the arithmetic to match any of the given options (0.1, 1, 10, 100 M) the intended quantity is 10 moles of solute (a common wording/printing slip in this recurring question, where 'kg' should read 'mol') - with no molar mass given, that is the only value that lets the problem be solved from the stated data. Taking the solute amount as 10 mol:

$M = \dfrac{10\ \text{mol}}{100\ \text{L}} = 0.1\ \text{mol/L}$

✓Final answer
$(a) 0.1 molar.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A solution contains 8 moles of solute and the mass of solvent is 4 kg. What is the molality of this solution?$
(a) 5 mol/kg
(b) 8 mol/kg
(c) 4 mol/kg
(d) 2 mol/kg
›Reveal solutionSolution
Molality is moles of solute per kilogram of solvent (not solution); here it works out to 8/4 = 2 mol/kg.

Molality (m) is defined as:

m = (moles of solute) / (mass of solvent in kg)

Given:

moles of solute = 8 mol

mass of solvent = 4 kg

m = 8 mol / 4 kg = 2 mol/kg

Note the common trap here: molality uses the mass of the SOLVENT, not the total solution mass or volume (that would be molarity or a different quantity) — a frequent point of confusion tested in this exact question style.

✓Final answer
$(d) 2 mol/kg.$
CBSE 2025Set ANNUAL1 mark
$Q. Define molality.$
›Reveal solutionSolution

Molality is defined as moles of solute per kilogram of solvent; unlike molarity, it does not depend on temperature since it is based on mass, not volume.

Molality (denoted m) of a solution is defined as the number of moles of solute dissolved in one kilogram (1000 g) of solvent:

Molality (m) = (number of moles of solute) / (mass of solvent in kg)

Unit: mol kg^-1 (also written as 'm', e.g., a '1 molal' or '1 m' solution).

Unlike molarity (which is based on the volume of the solution and therefore varies slightly with temperature, since volume expands/contracts with temperature), molality is based on the mass of the solvent, which does not change with temperature — so molality is a temperature-independent way of expressing concentration.

✓Final answer
$Molality is the number of moles of solute per kilogram of solvent: m = moles of solute / mass of solvent (in kg).$
CBSE 2024Set A11 mark
$Q.The number of moles of solute present in one kilogram of the solvent is called \rule{2cm}{0.4pt}.$
›Reveal solutionSolution

Moles of solute per kilogram of solvent defines molality.

Molality ( $m$ ) is a concentration term that depends only on the mass of solvent (and is therefore temperature-independent):

$m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} \;\; (\text{mol kg}^{-1})$

Since the definition specifies moles of solute in one kilogram of solvent, the quantity described is molality.

✓Final answer
$molality$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following has no unit?$
(a) Molarity
(b) Molality
(c) Normality
(d) Molar Fraction
›Reveal solutionSolution

Mole fraction is a dimensionless ratio, so it has no unit.

Molarity has units of $mol\,L^{-1}$ , molality has units of $mol\,kg^{-1}$ , and normality has units of $eq\,L^{-1}$ . Mole (molar) fraction is defined as the ratio of moles of one component to the total moles of all components in the solution, $x_i = \dfrac{n_i}{\sum n}$ — since it is a ratio of like quantities, it is a pure number and carries no unit.

✓Final answer
$(iv) Molar Fraction$
CBSE 2024Set ANNUAL1 mark
$Q. Write down the formula of Molarity.$
›Reveal solutionSolution

Molarity is the number of moles of solute dissolved per litre of solution.

Molarity is one of the most widely used units of concentration. It is defined as the number of moles of solute dissolved in one litre (one cubic decimetre) of solution:

$M = \frac{\text{moles of solute } (n)}{\text{volume of solution in litres } (V)}$

Equivalently, in terms of mass, $M = \dfrac{w \times 1000}{M_{r} \times V(mL)}$ , where $w$ is the mass of solute in grams and $M_r$ is its molar mass.

✓Final answer
$M = \dfrac{n}{V(L)}\ mol\,L^{-1}$

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Q.A measured temperature on Fahrenheit scale is 200 °F. What will this reading be on Celsius scale?

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?

Quick Comparison: Molarity vs Molality

Final Takeaway

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?

Common Mistake to Avoid

Quick Check: Why This Matters in Exams

Concept: Temperature Conversion Between Fahrenheit and Celsius

Method: Formula Substitution Method

Steps:

Mistake 1: Using the Wrong Formula (Inverting the Relationship)

Mistake 2: Forgetting to Subtract 32 First

Mistake 3: Incorrect Arithmetic with the Fraction $\frac{5}{9}$

Mistake 4: Confusing the Answer with a Nearby Trap Option

Summary Table for Quick Revision

Q.A measured temperature on Fahrenheit scale is 200 °F. What will this reading be on Celsius scale?

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