Chemistry — Class 11 practice questions

Worked examples for Chemistry — Class 11

1. Question 1

   The number of neutrons in the nuclide $^{80}_{35}\mathrm{Br}$ is

   1. A. $35$

      You reported the atomic number, which counts the protons in the nucleus (and the electrons of the neutral atom), not the neutrons.

   2. B. $80$

      You reported the mass number, which counts protons AND neutrons together; the protons must still be subtracted.

   3. C. $45$Correct answer

      The subscript is the atomic number and the superscript the mass number, so the neutrons number $A - Z = 80 - 35 = 45$.

   4. D. $115$

      You added the mass number and the atomic number; the nucleons are already included in the mass number, so the two are subtracted, never added.

   Explanation

   In a nuclide symbol the subscript is the atomic number $Z$, the number of protons, and the superscript is the mass number $A$, the total number of protons and neutrons in the nucleus. The neutron count is therefore $A - Z$, read straight off the symbol. Two nuclides with the same $A - Z$ are isotones, which is why this difference, and not $A$ or $Z$ alone, is the quantity the chapter asks for.

2. Question 2

   Hydrogen and oxygen diffuse through the same porous plug at the same temperature and pressure. Given:

   $M_{\mathrm{H_2}} = 2\ \text{g mol}^{-1}$

   $M_{\mathrm{O_2}} = 32\ \text{g mol}^{-1}$

   The ratio of the rate of diffusion of hydrogen to that of oxygen, $r_{\mathrm{H_2}} : r_{\mathrm{O_2}}$, is

   1. A. $16:1$

      You took the plain ratio of the molar masses, $32/2$, and left out the square root that the law carries.

   2. B. $4:1$Correct answer

      The rate of diffusion varies inversely as the square root of the molar mass, so $r_{\mathrm{H_2}}/r_{\mathrm{O_2}} = \sqrt{M_{\mathrm{O_2}}/M_{\mathrm{H_2}}} = \sqrt{32/2} = \sqrt{16} = 4$: the lighter gas diffuses four times faster.

   3. C. $1:4$

      You put the molar masses the wrong way up under the root, $\sqrt{2/32}$, which makes the lighter gas diffuse more slowly than the heavier one.

   4. D. $1:16$

      You used the masses the wrong way up and also dropped the square root, so hydrogen comes out sixteen times slower than oxygen.

   Explanation

   At one temperature every gas has the same average kinetic energy per molecule, so a lighter molecule must move faster; the molecular speed, and with it the rate of diffusion, goes as the inverse square root of the molar mass. Hence the ratio of rates is the square root of the inverse ratio of molar masses, and the lighter gas is always the faster one.

3. Question 3

   5 g of $\mathrm{CaCO_3}$ is treated with 7.3 g of HCl and the reaction $\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2}$ goes to completion.

   Given:

   $M_{\mathrm{CaCO_3}} = 100\ \text{g mol}^{-1}$

   $M_{\mathrm{HCl}} = 36.5\ \text{g mol}^{-1}$

   The molar volume of a gas at STP (273 K, 1 atm) is $22.4\ \text{dm}^{3},\text{mol}^{-1}$.

   The volume of $\mathrm{CO_2}$ collected at STP is

   1. A. $4.48\ \text{dm}^{3}$

      You carried the 0.2 mol of HCl straight across to 0.2 mol of $\mathrm{CO_2}$, as if every coefficient were one; the equation needs two moles of acid per mole of gas, and in any case the carbonate runs out first.

   2. B. $2.24\ \text{dm}^{3}$

      You made the acid the limiting reagent: 0.2 mol of HCl divided by its coefficient 2 gives 0.1 mol of $\mathrm{CO_2}$. The acid is in excess here — 0.05 mol of carbonate asks for only 0.1 mol of the 0.2 mol available — so it cannot fix the yield.

   3. C. $1.12\ \text{dm}^{3}$Correct answer

      $5/100 = 0.05$ mol of carbonate needs $2 \times 0.05 = 0.1$ mol of HCl, and $7.3/36.5 = 0.2$ mol is available, so the carbonate is limiting. It gives 0.05 mol of $\mathrm{CO_2}$, and $0.05 \times 22.4 = 1.12\ \text{dm}^{3}$ at STP.

   4. D. $2.2\ \text{g}$

      This is the mass of 0.05 mol of $\mathrm{CO_2}$, not its volume: the question asks for a gas volume, so the amount must be multiplied by the molar volume and not by the molar mass.

   Explanation

   When two reactant amounts are stated, the yield belongs to the reactant that runs out first, and that is decided by comparing available moles with the coefficients of the balanced equation — never by which mass is smaller or which reactant is named first. The carbonate here supplies 0.05 mol against an acid requirement it can meet twice over, so it is limiting and passes its amount to carbon dioxide in a 1:1 ratio. The molar volume, a convention tied to the stated standard conditions, then turns that amount into a volume.

4. Question 4

   The numbers of sigma and pi bonds present in propyne, $\mathrm{CH_3-C\equiv CH}$, are respectively

   1. A. $5$ and $3$

      You counted all three bonds of the triple bond as pi bonds. A triple bond is always one sigma and two pi, because the first overlap between two atoms is head-on.

   2. B. $2$ and $2$

      The bonds to hydrogen have been dropped. Each of the four C-H bonds is a sigma bond and must be counted along with the carbon-carbon bonds.

   3. C. $6$ and $0$

      You counted the triple bond as a single linkage. Six sigma bonds is right only if the two pi bonds of the triple bond are left out of the count.

   4. D. $6$ and $2$Correct answer

      Four C-H bonds and the C-C single bond give five sigma bonds, and the triple bond contributes one sigma and two pi: six sigma and two pi in all.

   Explanation

   Every covalent linkage between two atoms contains exactly one sigma bond, formed by head-on overlap; any further bonding between the same pair of atoms is pi, from sideways overlap of unhybridised p orbitals. So a single bond is one sigma, a double bond one sigma and one pi, and a triple bond one sigma and two pi, and the total is obtained by counting every bond in the structure, hydrogens included.

5. Question 5

   The molar mass of magnesium nitrate, $\mathrm{Mg(NO_3)_2}$, is

   Given: Mg = 24, N = 14, O = 16

   1. A. $100\ \text{g mol}^{-1}$

      You applied the subscript 2 to the nitrogen only and left the oxygen at three atoms: $24 + 2(14) + 3(16) = 100$. The bracket subscript multiplies every atom inside it.

   2. B. $86\ \text{g mol}^{-1}$

      You dropped the bracket and added up $\mathrm{MgNO_3}$ instead: $24 + 14 + 3(16) = 86$. The subscript 2 outside the bracket means the formula unit carries two whole nitrate groups.

   3. C. $134\ \text{g mol}^{-1}$

      You applied the subscript 2 to the oxygen only: $24 + 14 + 6(16) = 134$. The nitrogen is inside the same bracket, so it is doubled too.

   4. D. $148\ \text{g mol}^{-1}$Correct answer

      One nitrate group is $14 + 3(16) = 62$, and the subscript outside the bracket multiplies the whole group, so the formula unit is $24 + 2(62) = 148\ \text{g mol}^{-1}$.

   Explanation

   Molar mass is the sum of the atomic masses of every atom in one formula unit. A subscript written after a closing bracket multiplies each atom inside that bracket, so $\mathrm{Mg(NO_3)_2}$ contains one Mg, two N and six O atoms. Counting the atoms correctly before adding is the whole of the work.

6. Question 6

   Fog is made up of fine droplets of water dispersed in air. This colloidal system is classified as

   1. A. An aerosolCorrect answer

      The dispersed phase is a liquid and the dispersion medium is a gas; a liquid (or a solid) dispersed in a gas is an aerosol, the class to which fog, mist and clouds belong.

   2. B. A foam

      You have interchanged the phase and the medium. A foam is a gas dispersed in a liquid, as in soap lather or whipped cream.

   3. C. An emulsion

      An emulsion is a liquid dispersed in a liquid, as in milk; here the medium is air, a gas, not a liquid.

   4. D. A gel

      A gel is a liquid dispersed in a solid, as in cheese or jellies; the water droplets of fog are dispersed in a gas.

   Explanation

   A colloid is named from the physical states of its dispersed phase and its dispersion medium, in that order. Water droplets are the dispersed phase and air is the dispersion medium, so the system is a liquid in a gas, which the book classes as an aerosol. Reading the pair the other way round, or shifting one state, gives a different class of colloid altogether.

7. Question 7

   Cetyltrimethylammonium bromide, $\mathrm{CH_3(CH_2)_{15}N^{+}(CH_3)_3Br^{-}}$, is used in hair conditioners and as a germicide. It is classified as

   1. A. a cationic detergentCorrect answer

      Cationic detergents are quaternary ammonium salts carrying one long alkyl chain; the cleansing part of the molecule is the positively charged ion, exactly as here.

   2. B. an anionic detergent

      You classified the molecule by the bromide ion present. An anionic detergent is a sodium salt of a long-chain alkyl hydrogen sulfate or of an alkylbenzene sulfonic acid, where the long chain sits on the negative ion.

   3. C. a non-ionic detergent

      A non-ionic detergent carries no charge at all — it is an ester or an ether, such as a polyethylene glycol stearate; this compound is a salt and is fully ionic.

   4. D. a soap

      A soap is the sodium or potassium salt of a long-chain fatty acid. This has an ammonium centre and no carboxylate group, so it is a synthetic detergent, not a soap.

   Explanation

   Synthetic detergents are classified by the charge on the ion that does the cleaning. A quaternary ammonium salt with one long hydrocarbon chain puts that chain on the cation, so it is a cationic detergent; its germicidal action is why such compounds appear in hair conditioners and antiseptic preparations.

8. Question 8

   The colloidal particles of an arsenic sulphide ($\mathrm{As_2S_3}$) sol carry a negative charge. Which of the following ions is the most effective in coagulating this sol?

   1. A. $\mathrm{PO_4^{3-}}$

      You ranked by valency alone and chose an ion of the same sign as the sol particles; a negative ion is repelled by a negatively charged sol and cannot neutralise its charge, whatever its valency.

   2. B. $\mathrm{Ba^{2+}}$

      This cation does coagulate the sol, but it is divalent; its coagulating power is lower than that of a trivalent cation, so it is not the most effective ion in this set.

   3. C. $\mathrm{Na^{+}}$

      You picked an oppositely charged ion but ignored the valency part of the Hardy–Schulze rule; a monovalent cation has the least coagulating power of the three cations listed.

   4. D. $\mathrm{Al^{3+}}$Correct answer

      By the Hardy–Schulze rule a sol is coagulated by the ion of charge opposite to that of its particles, and the coagulating power rises with the valency of that ion. Among the cations offered, $\mathrm{Al^{3+}}$ carries the highest positive charge, so it is the most effective for a negatively charged sol.

   Explanation

   Coagulation happens when the charge on the colloidal particles is neutralised, so the effective ion must be the one of opposite sign — here a cation, because the sol is negative. The Hardy–Schulze rule then adds that the coagulating power of that ion increases with its valency, which makes a trivalent cation far more effective than a divalent or monovalent one. Both halves of the rule, the sign and the valency, have to be used.