Easy Practice — Set 1

Quick Recap — The Mole & Molar Mass

• Mole: $1$ mol $=6.022\times10^{23}$ particles (Avogadro number $N_A$).
• Moles from mass: $n=\dfrac{\text{mass}}{\text{molar mass}}$.
• Molar mass (g/mol) = sum of atomic masses. Use H=1, C=12, N=14, O=16, Na=23, S=32, Cl=35.5, Ca=40.
• Gas at STP: $1$ mol occupies $22.4$ L.
• Particles: number $=n\times N_A$.

Beyond-NCERT JEE Formulae

A rapid-reference sheet of the exam-only shortcuts that collapse multi-step mole problems into a single line. The basics (mole, molar mass, molarity, molality, dilution) are recapped in Sets 1 and 3 — this is the upgrade layer.

1. The mole — master conversion. Use whenever you must jump between mass, particle count and gas volume in one shot.

$$n=\dfrac{\text{mass}}{M}=\dfrac{N}{N_A}=\dfrac{V(\text{gas at STP})}{22.4\ \text{L}}$$

[JEE Tip] All three ratios equal the same n, so chain them directly (mass straight to molecules) instead of solving in stages. Classical STP uses 22.4 L per mole; if the paper states "1 bar STP", switch to 22.7 L.

2. Concentration terms — one for every situation.

• Molarity: $M=\dfrac{n_{solute}}{V_{soln}(\text{L})}$ — when the volume of solution is given.
• Molality: $m=\dfrac{n_{solute}}{\text{mass of solvent (kg)}}$ — temperature-independent, so it drives colligative-property questions.
• Mole fraction: $x_A=\dfrac{n_A}{n_A+n_B}$, with $x_A+x_B=1$.
• Normality: $N=M\times(\text{n-factor})$ — for acid-base and redox titrations.
• ppm: $\text{ppm}=\dfrac{\text{mass of solute}}{\text{mass of solution}}\times10^6$ — for trace (very dilute) solutions.
• Percentage strengths: %(w/w), %(w/v) and %(v/v).

[JEE Tip] Molarity falls when a solution is warmed (the volume expands); molality, mole fraction and mass % never change with temperature. If a question heats or cools a solution, that difference is usually the whole point.

3. Molarity to molality — the density bridge. Use when a solution is quoted as molarity plus density and you need molality (or the reverse).

$$m=\dfrac{1000\,M}{1000\,d-M\,M_2}$$

Here d is the density in g/mL and $M_2$ is the molar mass of the solute.

[JEE Tip] The product $M\,M_2$ is the mass of solute in one litre, so $1000\,d-M\,M_2$ is the mass of solvent in grams. Re-derive it once from 1 L of solution and you will never misremember the sign.

4. Molarity straight from a commercial bottle. Use for concentrated reagents labelled by %(w/w) and density.

$$M=\dfrac{10\times\%(w/w)\times d}{M_2}$$

[JEE Tip] The 10 is simply 1000 mL divided by 100%. Lab-bottle molarity of concentrated HCl, HNO3 or H2SO4 is a recurring one-mark item — memorise this line.

5. Empirical and molecular formula. Use for percentage-composition or combustion data.

$$MF=n\times EF,\qquad n=\dfrac{\text{molar mass}}{\text{EF mass}}$$

Divide each element's mass % by its atomic mass, then divide every result by the smallest to get the whole-number mole ratio (the empirical formula).

[JEE Tip] If a subscript lands on x.5, multiply all subscripts by 2 (x.33 by 3; x.25 by 4) rather than rounding — never round 2.5 up to 3.

6. Equivalent weight and n-factor. Use in normality and any equivalence (titration or redox) calculation.

$$\text{Equivalent weight}=\dfrac{M}{\text{n-factor}}$$

n-factor = basicity of an acid, acidity of a base, or electrons transferred per formula unit in a redox change.

[JEE Tip] At the end point of any titration, equivalents are equal: $N_1V_1=N_2V_2$. This is faster than balancing the redox equation and needs no mole ratio.

7. Limiting reagent — the yield gatekeeper. Use in every reaction that fixes the amounts of two reactants.

Convert each reactant to moles, divide by its stoichiometric coefficient, and the smallest quotient marks the limiting reagent — it alone fixes how much product forms.

[JEE Tip] Never compare raw masses or raw moles; always compare moles divided by coefficient. Leftover of the excess reagent = initial amount minus the amount actually consumed.

Solved Examples — Beyond-NCERT Formulae

Example 1: Molarity to molality using density. A 3.60 M aqueous solution of H2SO4 (molar mass 98) has a density of 1.19 g/mL. Find its molality.

Solution:

1. Apply the density bridge $m=\dfrac{1000\,M}{1000\,d-M\,M_2}$ with $M=3.60$, $d=1.19$ and $M_2=98$.
2. Numerator: $1000\times3.60=3600$.
3. Denominator (mass of solvent in 1 L, in grams): $1000\times1.19-3.60\times98=1190-352.8=837.2$.
4. $m=\dfrac{3600}{837.2}=4.30$ mol/kg.
5. Answer: molality = 4.30 m. It exceeds the molarity, as expected — the solvent mass (0.837 kg) is less than the 1 L of whole solution.

Example 2: Molarity of a concentrated acid. Concentrated nitric acid is 63% HNO3 by mass and has a density of 1.40 g/mL. Calculate its molarity (molar mass of HNO3 = 63).

Solution:

1. For a bottle labelled by %(w/w) and density, use $M=\dfrac{10\times\%(w/w)\times d}{M_2}$.
2. Substitute: $M=\dfrac{10\times63\times1.40}{63}$.
3. Numerator $=882$; divide by 63.
4. $M=14$ mol/L.
5. Answer: molarity = 14 M. The factor 10 converts the per-100 g basis to per-litre, so no sample size need be assumed.

Example 3: Molecular formula from percentage composition. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, with a molar mass of 180 g/mol. Find its molecular formula.

Solution:

1. Divide each mass % by the atomic mass: C $=\dfrac{40.0}{12}=3.33$, H $=\dfrac{6.7}{1}=6.70$, O $=\dfrac{53.3}{16}=3.33$.
2. Divide by the smallest (3.33): C $=1$, H $=2$, O $=1$, so the empirical formula is CH2O.
3. EF mass $=12+2(1)+16=30$.
4. $n=\dfrac{\text{molar mass}}{\text{EF mass}}=\dfrac{180}{30}=6$.
5. Multiply every subscript in CH2O by $n=6$.
6. Answer: C6H12O6 (glucose).

Example 4: Limiting reagent and yield. 28 g of N2 is mixed with 9 g of H2 and allowed to react: N2 + 3H2 -> 2NH3. Find the mass of NH3 formed and the mass of the excess reagent left over (N = 14, H = 1).

Solution:

1. Moles: N2 $=\dfrac{28}{28}=1$ mol; H2 $=\dfrac{9}{2}=4.5$ mol.
2. Divide by the coefficients: N2 gives $\dfrac{1}{1}=1$; H2 gives $\dfrac{4.5}{3}=1.5$. The smaller value is N2's, so N2 is the limiting reagent.
3. From the equation, 1 mol N2 makes 2 mol NH3 $=2\times17=34$ g.
4. H2 consumed $=3\times1=3$ mol $=6$ g, so H2 left $=9-6=3$ g.
5. Answer: 34 g of NH3 formed; 3 g of H2 remains unreacted.

Example 5: Normality in a titration. What volume of 0.5 N NaOH is required to exactly neutralise 25 mL of 0.2 M H2SO4?

Solution:

1. H2SO4 is diprotic, so its n-factor is 2 and $N=M\times(\text{n-factor})=0.2\times2=0.4$ N.
2. At the end point, equivalents are equal: $N_1V_1=N_2V_2$.
3. $0.4\times25=0.5\times V_2$, so $V_2=\dfrac{10}{0.5}$.
4. $V_2=20$ mL.
5. Answer: 20 mL of 0.5 N NaOH. Working in normality avoids balancing 2NaOH + H2SO4 -> Na2SO4 + 2H2O.

Example 6: Dilution of a stock solution. To what final volume must 100 mL of 2.0 M HCl be diluted to obtain a 0.25 M solution, and how much water is added?

Solution:

1. Moles of HCl stay fixed on dilution, so $M_1V_1=M_2V_2$.
2. $2.0\times100=0.25\times V_2$.
3. $V_2=\dfrac{200}{0.25}=800$ mL.
4. Water to add $=800-100=700$ mL.
5. Answer: dilute to 800 mL total, i.e. add 700 mL of water.

Example 7: Trace concentration in ppm. A 500 g water sample contains 0.005 g of dissolved oxygen. Express the concentration in parts per million.

Solution:

1. For dilute solutions, $\text{ppm}=\dfrac{\text{mass of solute}}{\text{mass of solution}}\times10^6$.
2. Mass of solution $\approx500$ g, since the 0.005 g of solute is negligible.
3. $\text{ppm}=\dfrac{0.005}{500}\times10^6$.
4. $=1\times10^{-5}\times10^6=10$.
5. Answer: 10 ppm.

Example 8: Combustion analysis to molecular formula. 0.60 g of an organic compound made only of C, H and O is burnt completely, giving 0.88 g of CO2 and 0.36 g of H2O. If its molar mass is 60 g/mol, find the molecular formula.

Solution:

1. Carbon: moles of CO2 $=\dfrac{0.88}{44}=0.02$, so mass of C $=0.02\times12=0.24$ g.
2. Hydrogen: moles of H2O $=\dfrac{0.36}{18}=0.02$; each water holds 2 H, so moles of H $=0.04$ and mass of H $=0.04$ g.
3. Oxygen (by difference): mass of O $=0.60-(0.24+0.04)=0.32$ g, so moles of O $=\dfrac{0.32}{16}=0.02$.
4. Mole ratio C : H : O $=0.02:0.04:0.02=1:2:1$, giving the empirical formula CH2O (EF mass 30).
5. $n=\dfrac{60}{30}=2$, so double every subscript in CH2O.
6. Answer: C2H4O2 (acetic acid).