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.

TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$0.892\,N$
›Reveal solutionSolution

" $20$ volume" means $1\ \text{L}$ of solution releases $20\ \text{L}$ of O $_2$ at STP; using $N=\dfrac{\text{volume strength}}{5.6}$ gives $N=\dfrac{20}{5.6}=3.57\ N$ - option (D).

Meaning of volume strength. A " $20$ volume" H $_2$ O $_2$ solution liberates $20\ \text{L}$ of O $_2$ (at STP) per litre of solution on decomposition:
$2\,\text{H}_2\text{O}_2\rightarrow2\,\text{H}_2\text{O}+\text{O}_2$
Step 1 - Moles of O $_2$ per litre.
$n_{\text{O}_2}=\frac{20}{22.4}=0.893\ \text{mol}$
Step 2 - Moles and equivalents of H $_2$ O $_2$ .

Two moles of H $_2$ O $_2$ give one mole of O $_2$ , so $n_{\text{H}_2\text{O}_2}=2\times0.893=1.786\ \text{mol L}^{-1}$ (molarity $=1.786\ M$ ). As an oxidant the $n$ -factor of H $_2$ O $_2$ is $2$ , so
$N=2\times M=2\times1.786=3.57\ N$
Shortcut (standard relation). Since $1\ N$ H $_2$ O $_2$ liberates $5.6\ \text{L}$ of O $_2$ per litre,
$N=\frac{\text{volume strength}}{5.6}=\frac{20}{5.6}=3.57\ N$
✓Final answer
$20$
TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$_3$
›Reveal solutionSolution
The key is to use the balanced chemical equation and stoichiometry to find the mass of pure HCl needed, then convert to the mass of the 50% w/w solution. The required mass is 292 g, so option (B) is correct.

Concept & Intuition

This problem is about reacting hydrochloric acid with calcium carbonate. The reaction is a classic acid–carbonate neutralization:
$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
We are given a 50% w/w solution — meaning 50 g of pure HCl per 100 g of solution. So the actual mass of solution needed will be double the mass of pure HCl required. The trap is forgetting to account for the dilution and just picking the mass of pure HCl.

Step-by-step solution

Write the balanced equation

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
This tells us: 1 mole of CaCO₃ reacts with 2 moles of HCl.

Find moles of CaCO₃ Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3×16 (O) = 100 g/mol. Given mass = 200 g.

$\text{Moles of CaCO}_3 = \frac{200}{100} = 2 \text{ mol}$

Find moles of HCl needed

From the equation: 1 mol CaCO₃ needs 2 mol HCl.

So 2 mol CaCO₃ need $2 \times 2 = 4$ mol HCl.

Find mass of pure HCl required

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.

$\text{Mass of pure HCl} = 4 \times 36.5 = 146 \text{ g}$

Convert to mass of 50% w/w solution A 50% w/w solution means: 50 g HCl in 100 g solution. So for 146 g pure HCl, the solution mass is:

$\text{Mass of solution} = 146 \times \frac{100}{50} = 146 \times 2 = 292 \text{ g}$
Watch out

A common mistake is to stop at 146 g (the mass of pure HCl) and pick option (C). But the question asks for the solution mass, not the pure acid mass.

Tip

For a 50% w/w solution, the mass of solution is always twice the mass of pure solute. So once you find pure HCl = 146 g, just double it to get 292 g.

✓Final answer
$The correct option is (B) . ANSWER: B$
TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
The "volume strength" of hydrogen peroxide is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution the strength (w/v) is about 15%, so the correct option is (D).

Concept & Intuition

"Volume strength" (e.g. "10 volume," "50 volume") labels a hydrogen peroxide solution by the volume of oxygen it releases: 1 mL of the solution produces that many mL of $\mathrm{O_2}$ at STP on decomposition. The decomposition is
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
To convert volume strength into a percentage (g of $\mathrm{H_2O_2}$ per 100 mL of solution), use the molar volume at STP (22.4 L/mol) and the molar mass of $\mathrm{H_2O_2}$ (34 g/mol).

Step-by-step reasoning

Interpret "50 volume"

1 mL of solution yields 50 mL of $\mathrm{O_2}$ at STP, so 1 L of solution yields $50 \times 1000 = 50\,000$ mL = 50 L of $\mathrm{O_2}$ .

Moles of $\mathrm{O_2}$

$n_{\mathrm{O_2}} = \frac{50}{22.4} \approx 2.232\ \text{mol}$

Moles of $\mathrm{H_2O_2}$

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times 2.232 = 4.464$ mol.

Mass of $\mathrm{H_2O_2}$ per litre

With molar mass 34 g/mol,

$m = 4.464 \times 34 \approx 151.8\ \text{g per litre}$

Express as a percentage (w/v) 151.8 g per 1000 mL means $\dfrac{151.8}{10} = 15.18$ g per 100 mL, i.e. approx 15%.

Tip

Shortcut: strength (w/v) $= \dfrac{\text{Volume strength} \times 34}{22.4 \times 10} \times 2 = \dfrac{\text{Volume strength} \times 68}{224}$ . For 50 volume: $\dfrac{50 \times 68}{224} \approx 15.18\%$ . The factor of 2 comes from the $2\!:\!1$ $\mathrm{H_2O_2}\!:\!\mathrm{O_2}$ ratio.

Watch out

A common mistake is to omit the 2:1 mole ratio between $\mathrm{H_2O_2}$ and $\mathrm{O_2}$ . Using 1:1 gives about 7.6%, which is wrong. Always start from the balanced equation.

✓Final answer
$The correct option is (D) . ANSWER: D$
TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
"Volume strength" of $\mathrm{H_2O_2}$ is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution, converting through the decomposition $2\mathrm{H_2O_2} \to 2\mathrm{H_2O} + \mathrm{O_2}$ gives a strength of about 15%, so the correct option is (D).

Why this approach works

"Volume strength" tells you how many millilitres of oxygen gas (at STP) one millilitre of the solution releases on decomposition. So a "50 volume" solution means 1 mL of solution yields 50 mL of $\mathrm{O_2}$ .

To convert this into a percentage by mass (w/v), we:

find the mass of $\mathrm{H_2O_2}$ that produces that volume of oxygen, using the decomposition stoichiometry;

relate that mass to the mass of solution (density approx 1 g/mL for dilute solutions).

The bridge is the balanced equation:
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
so 2 moles of $\mathrm{H_2O_2}$ give 1 mole of $\mathrm{O_2}$ .

Step-by-step reasoning

Moles of oxygen from the given volume At STP, 1 mole of gas occupies 22400 mL. For 50 mL of $\mathrm{O_2}$ (from 1 mL of solution):

$n_{\mathrm{O_2}} = \frac{50}{22400} = \frac{1}{448}\ \text{mol}$

Moles of $\mathrm{H_2O_2}$ that produce this oxygen

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times \frac{1}{448} = \frac{1}{224}\ \text{mol}$ .

Mass of $\mathrm{H_2O_2}$

Molar mass of $\mathrm{H_2O_2} = 34$ g/mol, so

$m = \frac{1}{224}\times 34 = \frac{34}{224} \approx 0.152\ \text{g}$

Express as a percentage 1 mL of solution has mass approx 1 g (density approx 1 g/mL), so

$\% = \frac{0.152}{1}\times 100 \approx 15.2\%$

Choose the closest option Among 50%, 25%, 10%, and 15%, the value 15.2% is nearest to 15%.

Tip

Handy shortcut: for an $X$ -volume $\mathrm{H_2O_2}$ solution, the strength (w/v) is $\% = \dfrac{17X}{56}$ . For $X = 50$ : $\dfrac{17 \times 50}{56} = \dfrac{850}{56} \approx 15.2\%$ .

Watch out

Do not forget the 2:1 mole ratio between $\mathrm{H_2O_2}$ and $\mathrm{O_2}$ . Using a 1:1 ratio gives about 7.6%, which is wrong. Always start from the balanced equation.

✓Final answer
$The correct option is (D) . ANSWER: D$

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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

Why this approach works

Step-by-step reasoning

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

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