Q.Calculate the molarity of each of the following solutions:

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The mole fraction of $\mathrm{CH_3OH}$ in an aqueous solution is 0.02. What is the molality of this solution? (A) 4.52 m (B) 3.39 m (C) 2.26 m (D) 1.13 m

   ›Reveal solutionSolution

   Convert mole fraction of solute directly to molality using $m=\dfrac{x_2}{x_1}\cdot\dfrac{1000}{M_1}$, taking $M_1=18\ \mathrm{g/mol}$ for water. Answer: 1.13 m.

   Concept and Intuition

   Molality ($m$ = moles solute per kg solvent) and mole fraction ($x_2$ = moles solute per total moles) are both intensive composition measures, so one converts to the other purely through mole-count and molar-mass arithmetic — no need to assume any solution volume or density, which is exactly why molality/mole-fraction problems are solvable without extra data (unlike molarity, which needs density).

   Step-by-Step Solution

   1. Given $x_2$ (mole fraction of $\mathrm{CH_3OH}$) $=0.02$, so $x_1$ (mole fraction of water) $=1-0.02=0.98$.
   2. Take a basis of 1 mole total solution: moles of $\mathrm{CH_3OH}$, $n_2=0.02$; moles of water, $n_1=0.98$.
   3. Mass of water (solvent) $= n_1 \times M_1 = 0.98 \times 18\ \mathrm{g/mol} = 17.64\ \mathrm{g} = 0.01764\ \mathrm{kg}$. …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The mole fraction of NaOH in aqueous NaOH solution is 0.02. What is the volume (in mL) of this solution that reacts completely with 1L of 0.5 M HCl solution? (density of water = 1 $g\ mL^{-1}$) (A) 220.5 (B) 661.5 (C) 441.3 (D) 882.6

   ›Reveal solutionSolution

   Find the NaOH needed to neutralize 0.5 mol HCl (= 0.5 mol NaOH), back out the water present at $x_{NaOH}=0.02$, and convert that water's mass to volume using its density — giving ≈441 mL.

   Concept and Intuition

   Mole fraction directly links the moles of solute to the moles of solvent present. Once we know exactly how many moles of NaOH must be present (fixed by the stoichiometric neutralization requirement), the mole-fraction relation pins down exactly how much water accompanies it — and hence, via water's known density, the volume of solution.

   Step-by-Step Solution

   1. Moles of HCl = $1\ L \times 0.5\ mol/L = 0.5$ mol.
   2. $NaOH + HCl \rightarrow NaCl + H_2O$ is 1:1, so moles of NaOH required = 0.5 mol.
   3. Mole fraction: $x_{NaOH} = \dfrac{n_{NaOH}}{n_{NaOH}+n_{H_2O}} = 0.02$.
   4. $\Rightarrow n_{NaOH}+n_{H_2O} = \dfrac{0.5}{0.02} = 25 \Rightarrow n_{H_2O} = 24.5$ mol.
   5. Mass of water = $24.5 \times 18\ g/mol \approx 441$ g. …

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3. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A solid solute is dissolved in water. The mole fraction of solute is 0.02. What is the molality of the solution? (A) 2.133 m (B) 2.5 m (C) 1.5 m (D) 1.133 m

   ›Reveal solutionSolution

   Convert mole fraction to molality by picking a convenient basis (1 mol total) and computing the solvent's mass. Answer: 1.133 m.

   Concept and Intuition

   Mole fraction and molality are related but different concentration scales: mole fraction is a ratio of moles, while molality is moles of solute per kilogram of solvent. To convert, assume a convenient total amount (1 mole of solution), find the moles of each component from the given mole fraction, convert the solvent's moles to a mass, and then compute molality directly.

   Step-by-Step Solution

   1. Given: mole fraction of solute $x_{solute}=0.02$, so mole fraction of water $x_{water}=1-0.02=0.98$.
   2. Assume 1 mol of total solution: $n_{solute}=0.02$ mol, $n_{water}=0.98$ mol.
   3. Mass of water $= n_{water}\times M_{water} = 0.98\times18 = 17.64$ g $= 0.01764$ kg. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.248 g of ethylene glycol ($\mathrm{C_2H_6O_2}$) is added to 200 g of water to prepare antifreeze. What is the molality of resultant solution? (C = 12 u; H = 1 u; O = 16 u) (A) 5 m (B) 10 m (C) 20 m (D) 40 m

   ›Reveal solutionSolution

   This is a direct molality calculation: moles of solute per kilogram of solvent.

   Concept and Intuition

   Molality is defined as moles of solute dissolved per kilogram of solvent (not solution), making it temperature-independent and ideal for colligative-property calculations like antifreeze formulations. Ethylene glycol's molar mass must first be computed from its formula to find the moles present.

   Step-by-Step Solution

   1. Molar mass of $C_2H_6O_2$: $2(12)+6(1)+2(16) = 24+6+32 = 62\ \mathrm{g\,mol^{-1}}$.
   2. Moles of ethylene glycol $= \dfrac{248}{62} = 4$ mol.
   3. Mass of water (solvent) $= 200\ \mathrm{g} = 0.200\ \mathrm{kg}$. …

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5. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.What is the approximate molality of 10% (w/w) aqueous glucose solution ? (Molar mass of glucose = 180 g mol$^{-1}$) (A) 0.31 m (B) 0.62 m (C) 0.93 m (D) 1.24 m

   ›Reveal solutionSolution

   Molality uses moles of solute per kg of solvent (not solution). For 10% w/w glucose, 100 g solution = 10 g glucose + 90 g water, giving molality $\approx 0.62$ m.

   Concept and Intuition

   Molality is defined relative to the mass of solvent only, unlike mass percent (w/w) which is defined relative to total solution mass. So the first step is always to extract the solvent mass from the given composition before applying $m = \dfrac{n_{solute}}{\text{mass of solvent (kg)}}$.

   Step-by-Step Solution

   1. Basis: 100 g of solution. 10% (w/w) glucose means 10 g glucose and $100-10=90$ g water.
   2. Moles of glucose $= \dfrac{10\text{ g}}{180\text{ g mol}^{-1}} = 0.05\overline{5}$ mol.
   3. Mass of solvent (water) in kg $= 90\text{ g} = 0.090$ kg. …

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6. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.A solution is prepared by adding 124 g of ethylene glycol (molar mass $= 62\ \mathrm{g\ mol^{-1}}$) to x g of water to get 10 m solution. What is the value of x (in g) ? (A) 100 (B) 400 (C) 800 (D) 200

   ›Reveal solutionSolution

   This is a direct molality calculation. 124 g of ethylene glycol is 2 mol; requiring a 10 m solution fixes the water mass at 200 g.

   Concept and Intuition

   Molality (m) is defined per kilogram of solvent, not solution — it is temperature-independent and depends only on moles of solute and mass of solvent.

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

   Step-by-Step Solution

   1. Moles of ethylene glycol $= \dfrac{124\ \text{g}}{62\ \text{g mol}^{-1}} = 2\ \text{mol}$.
   2. Let mass of water $= x$ g $= \dfrac{x}{1000}$ kg.
   3. Given molality $= 10$ m: $10 = \frac{2}{x/1000}$ …

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7. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.At 298 K, the density of an aqueous solution containing 82 g of acetic acid per $\mathrm{dm^3}$ is $1.01\ \mathrm{kg\,dm^{-3}}$. If the molarity of the solution is 'x' M, the molality (m) of the same solution is (molar mass of acetic acid $= 60\ \mathrm{g\,mol^{-1}}$) (A) $(1.856\,x)\ m$ (B) $(0.999\,x)\ m$ (C) $(0.928\,x)\ m$ (D) $(1.077\,x)\ m$

   ›Reveal solutionSolution

   Converting molarity to molality requires the mass of solvent (not solution), obtained by subtracting the solute's mass from the total solution mass computed via density.

   Concept and Intuition

   Molarity (mol/L of solution) and molality (mol/kg of solvent) are only related once you know the density of the solution, because you need to convert the solution's volume to solution mass and then subtract the solute mass to get solvent mass.

   Step-by-Step Solution

   1. Molarity: $x = \dfrac{82\ \mathrm{g}}{60\ \mathrm{g/mol}} = 1.3\overline{6}$ mol per litre of solution (so moles of acetic acid per litre solution $= x$).
   2. Mass of 1 L (1 dm³) of solution $= 1.01\ \mathrm{kg/dm^3} \times 1\ \mathrm{dm^3} = 1.01\ \mathrm{kg} = 1010$ g.
   3. Mass of solvent (water) $= 1010 - 82 = 928$ g $= 0.928$ kg. …

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8. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.1.06 g of $\mathrm{Na_2CO_3}$ (molar mass = 106 g $\mathrm{mol^{-1}}$) is dissolved in 500 g water. What is its molality? (A) 0.2 m (B) 0.02 m (C) 2 m (D) 0.04 m

   ›Reveal solutionSolution

   Direct molality calculation: moles of solute divided by kilograms of solvent gives 0.02 m.

   Concept and Intuition

   Molality is defined as moles of solute per kilogram of solvent (not solution), making it temperature-independent and convenient for colligative property calculations.

   Step-by-Step Solution

   1. Moles of $\mathrm{Na_2CO_3} = \dfrac{1.06\ \mathrm{g}}{106\ \mathrm{g\,mol^{-1}}} = 0.01$ mol.
   2. Mass of solvent (water) = 500 g = 0.500 kg.
   3. Molality $= \dfrac{0.01\ \mathrm{mol}}{0.500\ \mathrm{kg}} = 0.02$ mol/kg = 0.02 m. …

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9. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.The molarity of one molal glucose solution having density of 1.2 g/mL is (A) 0.101 M (B) 1.01 M (C) 2.01 M (D) 0.001 M

   ›Reveal solutionSolution

   Convert 1 molal glucose solution to molarity using the solution's density; molality and molarity differ because molarity is based on total solution volume, not solvent mass. Answer: 1.01 M.

   Concept and Intuition

   Molality (mol/kg solvent) and molarity (mol/L solution) are numerically close for dilute aqueous solutions but not identical, because molality ignores the solute's contribution to the total solution mass/volume. Density lets us convert between the two using $\text{Molarity} = \dfrac{1000 \times m \times d}{1000 + m\times M_w}$, where $m$ is molality, $d$ is density (g/mL), and $M_w$ is the solute's molar mass.

   Step-by-Step Solution

   1. Take 1 kg (1000 g) of water as solvent, containing 1 mol glucose ($M_w = 180\ \mathrm{g/mol}$, so mass = 180 g).
   2. Total solution mass $= 1000 + 180 = 1180$ g. …

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10. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.The molality of solution, when 18 g of glucose is added to the 18 g of $\mathrm{H_2O}$ is (A) 0.55 m (B) 2.55 m (C) 5.55 m (D) 55.5 m

    ›Reveal solutionSolution

    Molality is moles of solute per kilogram of solvent; with 0.1 mol glucose in 0.018 kg water, molality $=5.55\,\text{m}$.

    Concept and Intuition

    Molality is defined using the mass of solvent (not volume, and not total solution mass), which is exactly why it is temperature-independent and preferred in colligative-property calculations. Here both the solute (glucose) and the solvent (water) are given in grams, so both must first be converted appropriately — glucose to moles, water to kilograms.

    Step-by-Step Solution

    1. Moles of glucose ($M=180\,\text{g/mol}$): $n = \dfrac{18\,\text{g}}{180\,\text{g/mol}} = 0.1\,\text{mol}$.
    2. Mass of water in kg: $18\,\text{g} = 0.018\,\text{kg}$. …

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11. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.A sample of drinking water has 15 ppm (by mass) of a carcinogen (molar mass 120 g mol$^{-1}$). The molality of carcinogen in water sample in mol kg$^{-1}$ is (A) $2.50\times10^{-4}$ (B) $2.50\times10^{-3}$ (C) $1.25\times10^{-4}$ (D) $1.25\times10^{-3}$

    ›Reveal solutionSolution

    Converting 15 ppm (by mass) of a solute of molar mass 120 g/mol into molality gives $1.25\times10^{-4}$ mol kg$^{-1}$.

    Concept and Intuition

    ppm (parts per million) by mass means grams of solute per million grams of solution; for a dilute aqueous solution, the solution mass is essentially the water mass. Molality is moles of solute per kilogram of solvent, so we convert the mass basis to moles and then to per-kg terms.

    Step-by-Step Solution

    1. 15 ppm by mass $\Rightarrow$ 15 g of carcinogen per $10^6$ g ($=1000$ kg) of water.
    2. Moles of carcinogen $= \dfrac{15\text{ g}}{120\text{ g/mol}} = 0.125$ mol. …

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12. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
    Q.The molarity of 10% (w/w) aqueous NaOH solution (density 1.11 g $\mathrm{mL^{-1}}$) (A) 2.50 M (B) 3.25 M (C) 2.78 M (D) 1.52 M

    ›Reveal solutionSolution

    Converts a 10% w/w NaOH solution into molarity using its density and molar mass.

    Concept and Intuition

    Working with exactly 1 L (1000 mL) of solution makes the arithmetic simplest: the total mass of that litre comes from its density, 10% of that mass is NaOH, and dividing by NaOH's molar mass gives the number of moles present in that one litre — which is the molarity.

    Step-by-Step Solution

    1. Mass of 1 L (1000 mL) of solution $=1000\times1.11=1110$ g.
    2. Mass of NaOH in it (10% w/w) $=0.10\times1110=111$ g.
    3. Moles of NaOH $=\dfrac{111}{40}=2.775$ mol. …

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