JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 10

Multiple Choice Questions (MCQs) for JEE Chemistry: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: One atomic mass unit in kilogram is
(a) \( 1/N_A \)
(b) \( 12/N_A \)
(c) \( 1/(1000\,N_A) \)
(d) \( 1000/N_A \)
Show Answer & Explanation

Answer: (c) \( 1/(1000\,N_A) \)

Explanation:
1. One mole of carbon-12 weighs 12 g, so one atom weighs \( 12/N_A \) g.
2. 1 amu is 1/12 of this, that is \( 1/N_A \) g.
3. Converting grams to kilograms gives \( 1/(1000\,N_A) \) kg.

Teacher's Note:
1. 1 amu = 1/N_A gram. Do not forget the grams to kilograms conversion.

Question 2: The number of electrons in 3.1 mg of \( NO_3^- \) is (\( N_A = 6\times10^{23} \))
(a) 32
(b) \( 1.6\times10^{-3} \)
(c) \( 9.6\times10^{20} \)
(d) \( 9.6\times10^{23} \)
Show Answer & Explanation

Answer: (c) \( 9.6\times10^{20} \)

Explanation:
1. Molar mass of \( NO_3^- \) is 62 g, so 3.1 mg is \( 5\times10^{-5} \) mol.
2. Electrons per ion = 7 + 3(8) + 1 = 32.
3. Electrons = \( 5\times10^{-5}\times32\times6\times10^{23} = 9.6\times10^{20} \).

Teacher's Note:
1. Count electrons per ion, including the extra negative charge. Then multiply by the number of ions.

Question 3: A gaseous mixture contains \( CO_2 \)(g) and \( N_2O \)(g) in a 2 : 5 ratio by mass. The ratio of the number of molecules of \( CO_2 \)(g) and \( N_2O \)(g) is
(a) 5 : 2
(b) 2 : 5
(c) 1 : 2
(d) 5 : 4
Show Answer & Explanation

Answer: (b) 2 : 5

Explanation:
1. \( CO_2 \) and \( N_2O \) both have molar mass 44 g/mol.
2. Equal molar masses mean the mole ratio equals the mass ratio, 2 : 5.

Teacher's Note:
1. Whenever two gases have the same molar mass, mass ratio equals mole ratio.

Question 4: Which of the following contains the largest number of carbon atoms?
(a) 15 g ethane, \( C_2H_6 \)
(b) 40.2 g sodium oxalate, \( Na_2C_2O_4 \)
(c) 72 g glucose, \( C_6H_{12}O_6 \)
(d) 35 g pentene, \( C_5H_{10} \)
Show Answer & Explanation

Answer: (d) 35 g pentene, \( C_5H_{10} \)

Explanation:
1. Moles of C: (a) 15/30 x 2 = 1; (b) 40.2/134 x 2 = 0.6; (c) 72/180 x 6 = 2.4; (d) 35/70 x 5 = 2.5.
2. Pentene has the most.

Teacher's Note:
1. Convert to moles of the compound, then multiply by atoms per molecule.

Question 5: The number of hydrogen atoms in 0.9 g glucose, \( C_6H_{12}O_6 \), is the same as in
(a) 0.048 g hydrazine, \( N_2H_4 \)
(b) 0.17 g ammonia, \( NH_3 \)
(c) 0.30 g ethane, \( C_2H_6 \)
(d) 0.03 g hydrogen, \( H_2 \)
Show Answer & Explanation

Answer: (c) 0.30 g ethane, \( C_2H_6 \)

Explanation:
1. 0.9 g glucose is 0.005 mol, which holds 0.06 mol of H atoms.
2. Ethane: 0.30/30 = 0.01 mol x 6 = 0.06 mol H.
3. The others give 0.006, 0.03 and 0.03 mol.

Teacher's Note:
1. Compare mole of H atoms for each option. Only ethane matches.

Question 6: Ethanol, \( C_2H_5OH \), has density 0.7893 g/ml at 293 K. If 1.2 mol of ethanol is needed for an experiment, what volume should be measured out?
(a) 55 ml
(b) 58 ml
(c) 70 ml
(d) 79 ml
Show Answer & Explanation

Answer: (c) 70 ml

Explanation:
1. Mass = 1.2 x 46 = 55.2 g.
2. Volume = mass/density = 55.2/0.7893 = 69.9 ml, about 70 ml.

Teacher's Note:
1. Moles to mass using molar mass, then mass to volume using density.

Question 7: An iodized salt contains 0.5% of NaI. A person consumes 3 g of salt every day. The number of iodide ions going into his body every day is (I = 127)
(a) \( 10^{-4} \)
(b) \( 6.02\times10^{-4} \)
(c) \( 6.02\times10^{19} \)
(d) \( 6.02\times10^{23} \)
Show Answer & Explanation

Answer: (c) \( 6.02\times10^{19} \)

Explanation:
1. NaI in 3 g of salt = 0.015 g.
2. Molar mass of NaI = 150, so moles = \( 10^{-4} \).
3. Ions = \( 10^{-4}\times6.02\times10^{23} = 6.02\times10^{19} \).

Teacher's Note:
1. Each NaI gives one iodide ion. The answer is a number of ions, not moles.

Question 8: The percentage by mole of \( NO_2 \) in a mixture of \( NO_2 \)(g) and NO(g) having average molecular mass 34 is
(a) 25%
(b) 20%
(c) 40%
(d) 75%
Show Answer & Explanation

Answer: (a) 25%

Explanation:
1. Let the mole fraction of \( NO_2 \) be x. Then 46x + 30(1 - x) = 34.
2. 16x = 4, so x = 0.25, that is 25%.

Teacher's Note:
1. Average molar mass is the mole-fraction weighted mean of the molar masses.

Question 9: How many litres of oxygen at 1 atm and 273 K will be required to burn completely 2.2 g of propane (\( C_3H_8 \))?
(a) 11.2 L
(b) 22.4 L
(c) 5.6 L
(d) 44.8 L
Show Answer & Explanation

Answer: (c) 5.6 L

Explanation:
1. \( C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O \).
2. 2.2 g is 0.05 mol, which needs 0.25 mol \( O_2 \).
3. Volume = 0.25 x 22.4 = 5.6 L.

Teacher's Note:
1. Balance the equation first, then use 22.4 L per mole at 273 K and 1 atm.

Question 10: If 1/2 mole of oxygen combines with aluminium to form \( Al_2O_3 \), the weight of aluminium metal used in the reaction is (Al = 27)
(a) 27 g
(b) 18 g
(c) 54 g
(d) 40.5 g
Show Answer & Explanation

Answer: (b) 18 g

Explanation:
1. \( 4Al + 3O_2 \to 2Al_2O_3 \).
2. Moles of Al = 4/3 x 0.5 = 2/3 mol.
3. Mass = 2/3 x 27 = 18 g.

Teacher's Note:
1. The mole ratio Al : O2 is 4 : 3 from the balanced equation.

Question 11: The volume of \( O_2 \) obtained at 2 atm and 546 K by the complete decomposition of 8.5 g \( NaNO_3 \) is \( 2NaNO_3 \to 2NaNO_2 + O_2 \)
(a) 2.24 lit
(b) 1.12 lit
(c) 0.84 lit
(d) 0.56 lit
Show Answer & Explanation

Answer: (b) 1.12 lit

Explanation:
1. 8.5 g \( NaNO_3 \) is 0.1 mol, which gives 0.05 mol \( O_2 \).
2. \( V = nRT/P = 0.05\times0.0821\times546/2 = 1.12 \) L.

Teacher's Note:
1. The gas is not at STP, so use the ideal gas equation.

Question 12: The maximum mass of sucrose \( C_{12}H_{22}O_{11} \) produced by mixing 84 g of carbon, 12 g of hydrogen and 56 L of \( O_2 \) at 1 atm and 273 K, according to \( 12C + 11H_2 + \tfrac{11}{2}O_2 \to C_{12}H_{22}O_{11} \), is
(a) 138.5 g
(b) 155.5 g
(c) 172.5 g
(d) 199.5 g
Show Answer & Explanation

Answer: (b) 155.5 g

Explanation:
1. Moles available: C = 7, \( H_2 \) = 6, \( O_2 \) = 2.5.
2. Moles of sucrose possible: 7/12 = 0.583; 6/11 = 0.545; 2.5/5.5 = 0.4545.
3. Oxygen is limiting, so mass = 0.4545 x 342 = 155.5 g.

Teacher's Note:
1. Divide each moles by its coefficient. The smallest value shows the limiting reagent.

Question 13: The minimum mass of a mixture of \( A_2 \) and \( B_4 \) required to produce at least 1 kg of each product is (At. mass of A = 10, B = 120): \( 5A_2 + 2B_4 \to 2AB_2 + 4A_2B \)
(a) 2120 g
(b) 1060 g
(c) 560 g
(d) 1660 g
Show Answer & Explanation

Answer: (a) 2120 g

Explanation:
1. Masses per reaction: 5 \( A_2 \) = 100 g and 2 \( B_4 \) = 960 g, total 1060 g.
2. Products: 2 \( AB_2 \) = 500 g and 4 \( A_2B \) = 560 g.
3. To get at least 1 kg of \( AB_2 \) the reaction must be scaled by 2, which gives 1000 g and 1120 g. Reactant mass = 2120 g.

Teacher's Note:
1. The product with the smaller yield per batch decides the scale-up factor.

Question 14: The mass of \( CO_2 \) produced from 620 g of a mixture of \( C_2H_4O_2 \) and \( O_2 \), prepared to produce the maximum energy, is (the combustion reaction is exothermic)
(a) 413.33 g
(b) 593.04 g
(c) 440 g
(d) 320 g
Show Answer & Explanation

Answer: (c) 440 g

Explanation:
1. Maximum energy needs complete reaction in the stoichiometric ratio.
2. \( C_2H_4O_2 + 2O_2 \to 2CO_2 + 2H_2O \): 60 g + 64 g = 124 g of mixture.
3. 620/124 = 5, so \( CO_2 \) = 5 x 2 x 44 = 440 g.

Teacher's Note:
1. The phrase maximum energy means both reactants are used up completely.

Question 15: The mass of \( P_4O_{10} \) produced if 440 g of \( P_4S_3 \) is mixed with 384 g of \( O_2 \) is \( P_4S_3 + 8O_2 \to P_4O_{10} + 3SO_2 \)
(a) 568 g
(b) 426 g
(c) 284 g
(d) 396 g
Show Answer & Explanation

Answer: (b) 426 g

Explanation:
1. \( P_4S_3 \) = 220 g/mol, so 440 g is 2 mol. \( O_2 \) = 384/32 = 12 mol.
2. 2 mol \( P_4S_3 \) needs 16 mol \( O_2 \), so \( O_2 \) is limiting.
3. \( P_4O_{10} \) = 12/8 = 1.5 mol x 284 = 426 g.

Teacher's Note:
1. Balance first. Find the limiting reagent before computing the product.

Question 16: An ideal gaseous mixture of ethane (\( C_2H_6 \)) and ethene (\( C_2H_4 \)) occupies 28 litre at 1 atm and 0 degrees C. The mixture reacts completely with 128 g \( O_2 \) to produce \( CO_2 \) and \( H_2O \). The mole fraction of \( C_2H_6 \) in the mixture is
(a) 0.6
(b) 0.4
(c) 0.5
(d) 0.8
Show Answer & Explanation

Answer: (b) 0.4

Explanation:
1. Total gas = 28/22.4 = 1.25 mol; \( O_2 \) = 4 mol.
2. Let ethane be x mol: it needs 3.5x mol \( O_2 \) and ethene needs 3(1.25 - x).
3. 3.5x + 3.75 - 3x = 4, so x = 0.5.
4. Mole fraction = 0.5/1.25 = 0.4.

Teacher's Note:
1. Write an oxygen balance with one unknown.

Question 17: A mixture of \( Li_2CO_3 \) and \( Na_2CO_3 \) is heated strongly in an open vessel. If the loss in mass of the mixture is \( \frac{220}{9} \)%, the molar ratio of \( Li_2CO_3 \) and \( Na_2CO_3 \) in the initial mixture is (Li = 7, Na = 23)
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 11 : 4
Show Answer & Explanation

Answer: (a) 1 : 1

Explanation:
1. Only \( Li_2CO_3 \) (74 g/mol) decomposes, losing 44 g of \( CO_2 \) per mole. \( Na_2CO_3 \) (106) is stable.
2. 44x/(74x + 106y) = 220/900.
3. 39600x = 16280x + 23320y, so x = y. The ratio is 1 : 1.

Teacher's Note:
1. Sodium carbonate does not decompose on heating; lithium carbonate does.

Question 18: Calculate the percentage change in \( M_{avg} \) of the mixture if \( PCl_5 \) undergoes 50% decomposition in a closed vessel: \( PCl_5 \to PCl_3 + Cl_2 \)
(a) 50%
(b) 66.66%
(c) 33.33%
(d) Zero
Show Answer & Explanation

Answer: (c) 33.33%

Explanation:
1. Start with 1 mol \( PCl_5 \) (208.5 g). After 50% decomposition total moles = 1.5.
2. \( M_{avg} \) = 208.5/1.5 = 139.
3. Change = (208.5 - 139)/208.5 = 33.33% decrease.

Teacher's Note:
1. Total mass is conserved while the number of moles changes, so the average molar mass changes inversely.

Question 19: A metal carbonate decomposes as \( M_2CO_3 \)(s) \( \to M_2O \)(s) + \( CO_2 \)(g). The percentage loss in mass on complete decomposition of \( M_2CO_3 \) (atomic mass of M = 102) is
(a) \( \frac{100}{3} \)%
(b) \( \frac{50}{3} \)%
(c) \( \frac{25}{3} \)%
(d) 15%
Show Answer & Explanation

Answer: (b) \( \frac{50}{3} \)%

Explanation:
1. Molar mass of \( M_2CO_3 \) = 204 + 60 = 264.
2. Mass lost = 44 g of \( CO_2 \).
3. Loss = 44/264 x 100 = 50/3 %.

Teacher's Note:
1. The mass lost is just the CO2 given off.

Question 20: 90 g of a mixture of \( H_2 \) and \( O_2 \) is taken in the stoichiometric ratio and gives \( H_2O \) with 50% yield. The mass of \( H_2O \) produced (in g) is
(a) 45
(b) 36
(c) 20
(d) 90
Show Answer & Explanation

Answer: (a) 45

Explanation:
1. All 90 g of reactants would form 90 g of water by mass conservation if the yield were 100%.
2. At 50% yield the mass of water is 45 g.

Teacher's Note:
1. In a stoichiometric mixture with a single product, product mass equals total reactant mass.

Question 21: An impure sample of \( CaCO_3 \) contains 38% of Ca. The percentage of impurity present in the sample is
(a) 5%
(b) 95%
(c) 10%
(d) 2.5%
Show Answer & Explanation

Answer: (a) 5%

Explanation:
1. Pure \( CaCO_3 \) has 40% Ca.
2. Purity = 38/40 = 95%.
3. Impurity = 5%.

Teacher's Note:
1. Compare the observed percentage of an element with its percentage in the pure compound.

Question 22: A sample of \( NH_3 \) gas is 20% dissociated into \( N_2 \) and \( H_2 \) gases. The mass ratio of \( N_2 \) and \( NH_3 \) gases in the final sample is
(a) \( \frac{7}{34} \)
(b) \( \frac{7}{17} \)
(c) \( \frac{14}{17} \)
(d) \( \frac{21}{17} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{34} \)

Explanation:
1. Start with 1 mol \( NH_3 \). 0.2 mol dissociates giving 0.1 mol \( N_2 \) and 0.3 mol \( H_2 \).
2. Left: 0.8 mol \( NH_3 \) = 13.6 g and 0.1 mol \( N_2 \) = 2.8 g.
3. Ratio = 2.8/13.6 = 7/34.

Teacher's Note:
1. Convert moles to mass before taking a mass ratio.

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