Q.A sample of drinking water was found to be severely contaminated with chloroform ($CHCl_3$) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

Showing the 12 most recent of 28 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Chlorophyll contains 2.4% of magnesium. The number of magnesium atoms present in 2.0 g of chlorophyll is ($N = 6 \times 10^{23}\ mol^{-1}$, At.wt of Mg = 24 $u$) (A) $1.8 \times 10^{21}$ (B) $2.4 \times 10^{21}$ (C) $1.2 \times 10^{21}$ (D) $3.6 \times 10^{21}$

   ›Reveal solutionSolution

   A straightforward percentage-composition → moles → atoms calculation; the answer is $1.2\times10^{21}$ Mg atoms.

   Concept and Intuition

   Percentage composition tells us the mass fraction of an element in a compound. Once we know the mass of that element, dividing by its atomic mass gives moles, and multiplying by Avogadro's number gives the actual atom count — the standard mass→mole→number chain.

   Step-by-Step Solution

   1. Mass of Mg in 2.0 g chlorophyll $= 2.4\% \times 2.0\ g = 0.024 \times 2.0 = 0.048$ g.
   2. Moles of Mg $= \dfrac{0.048}{24} = 0.002$ mol. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The molar mass of mustard gas is 159 $g\,mol^{-1}$. The percentage by mass of sulphur and chlorine in it are respectively (Atomic weight: $Cl = 35.5\ u$, $S = 32\ u$) (A) 20.12, 44.65 (B) 44.65, 20.12 (C) 40, 24.65 (D) 24.65, 40

   ›Reveal solutionSolution

   Mustard gas, $(ClCH_2CH_2)_2S$ ($C_4H_8Cl_2S$, $M=159$), is 20.1% S and 44.65% Cl by mass.

   Concept and Intuition

   Percentage composition just needs the correct molecular formula and the atomic masses of the elements asked about, divided by the total molar mass.

   Step-by-Step Solution

   1. Mustard gas is bis(2-chloroethyl) sulfide, $(ClCH_2CH_2)_2S$, i.e. $C_4H_8Cl_2S$.
   2. Molar mass check: $C_4=48$, $H_8=8$, $Cl_2=2\times35.5=71$, $S=32$. Sum $=48+8+71+32=159\,g\,mol^{-1}$ — matches the given value, confirming the formula.
   3. Mass fraction of S $=\dfrac{32}{159}=0.2013 \Rightarrow 20.12\%$ (rounded as in the option).
   4. Mass fraction of Cl $=\dfrac{71}{159}=0.4465 \Rightarrow 44.65\%$.
   5. So %S, %Cl (respectively) $=20.12,\ 44.65$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A gas mixture contains 64% methane and 36% ethane by mass. The density of the mixture at $27°C$ and 750 mm pressure (in $g\ L^{-1}$) is (at. wt $C = 12\ u, H = 1\ u, R = 0.082\ L\ atm\ K^{-1}mol^{-1}$) (A) 0.57 (B) 0.87 (C) 0.67 (D) 0.77

   ›Reveal solutionSolution

   Compute the mixture's average molar mass from the mass percentages, then apply the ideal-gas density formula $d=PM/RT$ to get $\approx 0.77\ \text{g L}^{-1}$.

   Concept and Intuition

   For a gas mixture, the effective (average) molar mass is the total mass divided by total moles of all species combined. Once you have that average $M$, the density of the mixture behaves just like a single ideal gas of that molar mass: $d = PM/RT$.

   Step-by-Step Solution

   1. Assume 100 g of mixture: mass of $\mathrm{CH_4} = 64$ g, mass of $\mathrm{C_2H_6} = 36$ g.
   2. Moles of $\mathrm{CH_4}$ ($M=16$ g/mol) $= 64/16 = 4$ mol.
   3. Moles of $\mathrm{C_2H_6}$ ($M=30$ g/mol) $= 36/30 = 1.2$ mol.
   4. Total moles $= 4+1.2 = 5.2$ mol; total mass $=100$ g.
   5. Average molar mass $M_{avg} = 100/5.2 = 19.23$ g/mol.
   6. Convert pressure: $P = 750/760 = 0.9868$ atm; $T = 27°C = 300$ K. …

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4. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.At T(K), a gaseous mixture contains $\mathrm{H_2}$ and $\mathrm{O_2}$. The total pressure of the mixture is 2 bar. The partial pressure of $\mathrm{H_2}$ is 1.778 bar. What is the weight (w/w) percentage of $\mathrm{H_2}$ in the mixture ? (A) 66.67 (B) 33.33 (C) 80.00 (D) 20.00

   ›Reveal solutionSolution

   Tests converting partial-pressure (mole fraction) data into a weight percentage using molar masses; the answer is 33.33%.

   Concept and Intuition

   Partial pressure is proportional to mole fraction (Dalton's law: $p_i = x_i P_{total}$ at constant T, V). Once we know the mole fractions of each gas, we can find the mass of each component (for an assumed total of 1 mole of gas mixture) using their molar masses, and hence the weight percentage.

   Step-by-Step Solution

   1. Mole fraction of $H_2$: $x_{H_2} = \dfrac{p_{H_2}}{P_{total}} = \dfrac{1.778}{2} = 0.889$.
   2. Mole fraction of $O_2$: $x_{O_2} = 1 - 0.889 = 0.111$.
   3. Assume 1 total mole of gas mixture. Mass of $H_2 = 0.889 \times 2\ \text{g/mol} = 1.778$ g. Mass of $O_2 = 0.111 \times 32\ \text{g/mol} = 3.552$ g.
   4. Total mass = $1.778 + 3.552 = 5.33$ g. …

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5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.1.84 g of a mixture of $\mathrm{CaCO_3}$ and $\mathrm{MgCO_3}$ is strongly heated to get a residue of 0.96 g. The percentage of $\mathrm{CaCO_3}$ in the mixture is (A) 50.34 (B) 49.66 (C) 54.34 (D) 45.66

   ›Reveal solutionSolution

   Tests mass-balance on a two-component thermal decomposition mixture; solving simultaneous equations gives 54.34% $CaCO_3$.

   Concept and Intuition

   Both carbonates decompose on strong heating: $CaCO_3 \to CaO + CO_2$ and $MgCO_3 \to MgO + CO_2$, each losing $CO_2$ mass. Since the two salts lose mass in different fixed ratios (dictated by their molar masses), the residue mass depends on the mixture's composition — letting us solve for the unknown split.

   Step-by-Step Solution

   1. Let mass of $CaCO_3 = x$ g, so mass of $MgCO_3 = (1.84-x)$ g.
   2. $CaCO_3$ (M = 100 g/mol) → $CaO$ (M = 56 g/mol): mass of CaO produced $= \dfrac{56}{100}x = 0.56x$.
   3. $MgCO_3$ (M = 84 g/mol) → $MgO$ (M = 40 g/mol): mass of MgO produced $= \dfrac{40}{84}(1.84-x)$.
   4. Total residue: $0.56x + \dfrac{40}{84}(1.84-x) = 0.96$.
   5. Compute $\dfrac{40}{84} = 0.4762$: $0.56x + 0.4762(1.84) - 0.4762x = 0.96$ …

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6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The mole fractions of glucose and water in aqueous glucose solution are 0.0244 and 0.9756 respectively. What is the weight percentage (w/w) of glucose in this solution ? (A) 40 (B) 25 (C) 20 (D) 10

   ›Reveal solutionSolution

   Converting mole fraction to weight percentage for a glucose–water solution gives approximately 20% w/w glucose.

   Concept and Intuition

   Mole fraction tells us the mole ratio of components; to get weight percentage we must convert moles to mass using the molar masses (glucose $M = 180\ \mathrm{g/mol}$, water $M = 18\ \mathrm{g/mol}$), then take the mass fraction.

   Step-by-Step Solution

   1. Assume a total of 1 mole of solution (mole fractions given directly as fractions of 1 mole total).
   2. Moles of glucose $= 0.0244$, moles of water $= 0.9756$.
   3. Mass of glucose $= 0.0244 \times 180 = 4.392\ \mathrm{g}$.
   4. Mass of water $= 0.9756 \times 18 = 17.5608\ \mathrm{g}$.
   5. Total mass of solution $= 4.392 + 17.5608 = 21.9528\ \mathrm{g}$. …

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7. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.At T(K), a gaseous mixture contains $\mathrm{H_2}$ and $\mathrm{O_2}$. The total pressure of the mixture is 2 bar. The weight percentage (w/w) of $\mathrm{H_2}$ is 33.33%. What is the approximate ratio of partial pressure of $\mathrm{H_2}$ and $\mathrm{O_2}$? (A) 8 : 1 (B) 4 : 1 (C) 2 : 1 (D) 3 : 1

   ›Reveal solutionSolution

   Converting the given weight percentage to moles (using molar masses 2 and 32 g/mol) gives a partial pressure ratio of $8:1$.

   Concept and Intuition

   By Dalton's law of partial pressures, for a gas mixture at a common temperature and volume, the partial pressure of each component is proportional to its mole fraction (equivalently, to its number of moles). So converting mass percentages to moles (using each gas's molar mass) directly gives the pressure ratio.

   Step-by-Step Solution

   1. Take a convenient total mass, say $3\ \text{g}$, so that $33.33\%=1/3$ works out to whole numbers: mass of H$_2$ $=1\ \text{g}$, mass of O$_2$ $=2\ \text{g}$.
   2. Moles of H$_2$: $n_{H_2} = \frac{1}{2} = 0.5\ \text{mol}$
   3. Moles of O$_2$: $n_{O_2} = \frac{2}{32} = 0.0625\ \text{mol}$ …

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8. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.What are the mole fractions of glucose and water respectively, in 20% (w/w) aqueous glucose solution ? (C = 12 u; O = 16 u; H = 1 u) (A) 0.0244, 0.9756 (B) 0.04, 0.96 (C) 0.0636, 0.9364 (D) 0.0124, 0.9876

   ›Reveal solutionSolution

   Taking a 100 g basis of the 20% w/w solution gives 20 g glucose and 80 g water; converting to moles and dividing gives mole fractions 0.0244 (glucose) and 0.9756 (water).

   Concept and Intuition

   "20% w/w" means 20 g of solute per 100 g of solution, so the remaining 80 g is solvent. Mole fraction of each component is its moles divided by the total moles of all components.

   Step-by-Step Solution

   1. Basis: 100 g solution → 20 g glucose ($\mathrm{C_6H_{12}O_6}$, $M=180$ g/mol), 80 g water ($M=18$ g/mol).
   2. Moles of glucose $= 20/180 = 0.1111$ mol.
   3. Moles of water $= 80/18 = 4.4444$ mol.
   4. Total moles $= 0.1111 + 4.4444 = 4.5556$ mol.
   5. Mole fraction of glucose $= 0.1111/4.5556 = 0.0244$. …

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9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.In aqueous glucose solution, the mole fraction of water is 40 times to mole fraction of glucose. What is the weight percentage (w/w) of glucose in the solution? (A) 40 (B) 30 (C) 20 (D) 10

   ›Reveal solutionSolution

   This tests converting mole-fraction ratio into moles, then into a weight percentage. The answer is 20%.

   Concept and Intuition

   Mole fraction compares moles of a component to total moles, while weight percentage compares mass of a component to total mass. To go from one to the other we must first fix actual mole amounts (using the given ratio), then convert moles to grams using molar masses (glucose $M=180\ \mathrm{g\,mol^{-1}}$, water $M=18\ \mathrm{g\,mol^{-1}}$).

   Step-by-Step Solution

   1. Let mole fraction of glucose $=x_2$ and of water $=x_1$. Given $x_1 = 40x_2$.
   2. Since $x_1+x_2=1$: $40x_2+x_2=1 \Rightarrow x_2=\dfrac{1}{41},\ x_1=\dfrac{40}{41}$.
   3. Take $n_2=1$ mol glucose $\Rightarrow n_1=40$ mol water (same ratio as mole fractions since total moles cancel).
   4. Mass of glucose $=1\times180=180$ g. Mass of water $=40\times18=720$ g. …

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10. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.The mass of a mixture containing $\mathrm{NaCl}$ and $\mathrm{NaBr}$ is 4.0 g. If Na is 30% of the total mixture, the composition of $\mathrm{NaCl}$ in the mixture is (Na = 23 u, Cl = 35.5 u, Br = 80 u) (A) 48% (B) 55% (C) 45% (D) 52%

    ›Reveal solutionSolution

    This is a mixture stoichiometry problem: using the given %Na by mass to set up and solve a linear equation for the NaCl fraction, giving 45%.

    Concept and Intuition

    Each mole of NaCl and each mole of NaBr contributes exactly one mole of Na. So if we know the total mass of Na present, and express the unknown masses of NaCl and NaBr in terms of one variable, we can solve for the composition directly using molar masses.

    Step-by-Step Solution

    1. Total mixture mass = 4.0 g. Na is 30% of this by mass: mass of Na $= 0.30 \times 4.0 = 1.2$ g.
    2. Moles of Na $= \dfrac{1.2}{23} = 0.052174$ mol.
    3. Let mass of NaCl $= x$ g, so mass of NaBr $= (4.0 - x)$ g.
    4. Molar mass of NaCl $= 23+35.5 = 58.5$; molar mass of NaBr $= 23+80=103$.
    5. Moles of Na from both salts: $\dfrac{x}{58.5} + \dfrac{4.0-x}{103} = 0.052174$.
    6. Multiply through by $58.5\times103 = 6025.5$: $103x + 58.5(4.0-x) = 314.35$. …

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11. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.At 293 K the density of an aqueous solution containing 120 g of urea ($\mathrm{NH_2CONH_2}$) per $\mathrm{dm^3}$ is 1.02 kg $\mathrm{dm^{-3}}$. The mole fraction of urea is approximately (A) 0.038 (B) 0.962 (C) 0.02 (D) 0.98

    ›Reveal solutionSolution

    Using the solution's density to get the total mass per litre, subtracting the urea mass gives the water mass; converting both to moles gives a urea mole fraction of about 0.038.

    Concept and Intuition

    Mole fraction requires the moles of each component in a defined amount of solution. Density lets us convert the given volume (1 dm³ = 1 L) into total solution mass, from which subtracting the known solute mass gives the solvent (water) mass.

    Step-by-Step Solution

    1. Total solution mass in 1 dm³ = density × volume = $1.02\ \mathrm{kg/dm^3}\times1\ \mathrm{dm^3} = 1020$ g.
    2. Mass of urea = 120 g (given), so mass of water = $1020-120 = 900$ g.
    3. Molar mass of urea, $\mathrm{NH_2CONH_2}$: $2(14)+4(1)+12+16 = 28+4+12+16=60$ g/mol.
    4. Moles of urea $= 120/60 = 2$ mol.
    5. Moles of water $= 900/18 = 50$ mol. …

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12. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.The concentration of 1L of CaCO$_3$ solution is 1000 ppm. What is its concentration in mol L$^{-1}$? (Ca = 40 u, O = 16 u, C = 12 u) (A) $10^{-3}$ (B) $10^{-1}$ (C) $10^{-4}$ (D) $10^{-2}$

    ›Reveal solutionSolution

    Converting 1000 ppm of CaCO$_3$ (≈ 1 g/L for a dilute aqueous solution) into molarity using M = 100 g/mol gives $10^{-2}$ mol/L — option (D).

    Concept and Intuition

    "ppm" (parts per million) for a dilute aqueous solution is conventionally taken as mg of solute per litre of solution (since the solution's density is essentially that of water, 1 g/mL, at these very low concentrations). Once we have the mass concentration in g/L, dividing by the molar mass converts it to molarity directly.

    Step-by-Step Solution

    1. 1000 ppm = 1000 mg solute per L of solution = 1 g/L (for a dilute aqueous solution where density ≈ 1 g/mL).
    2. Molar mass of $\mathrm{CaCO_3}$: $40 (\text{Ca}) + 12 (\text{C}) + 3\times16 (\text{O}) = 40+12+48 = 100$ g/mol. …

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