JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 39

Answer: (c) 0.8

Explanation:
1. Density of water is 1 g/mL, so 1.8 mL has mass 1.8 g, which is 1.8/18 = 0.1 mol of water.
2. One \( \ce{H2O \)} molecule has 8 neutrons (hydrogen-1 has none, oxygen-16 has 8).
3. Moles of neutrons = 0.1 x 8 = 0.8.

Teacher's Note:
1. Count the particles per molecule first, then multiply by moles.
2. Common slip: treating 1.8 mL as 1.8 mol.

Answer: (a) \( \dfrac{16}{N_A \)} g

Explanation:
1. At 0°C and 2 atm, 350 mL of gas corresponds to 0.7 L at 1 atm (STP), which is 0.7/22.4 = 1/32 mol.
2. Molar mass = 1 g / (1/32 mol) = 32 g/mol for the diatomic molecule, so the atomic mass is 16 g/mol.
3. Mass of one atom = \( 16/N_A \) g.

Teacher's Note:
1. Convert the gas volume to STP before using 22.4 L/mol.
2. Mass of one atom = molar mass of the atom divided by Avogadro's number.

Answer: (b) \( \ce{C2H6 \)}

Explanation:
1. Let the hydrocarbon be \( \ce{C_xH_y \)}. Gas volumes are proportional to moles, so 10 mL gives 20 mL \( \ce{CO2 \)}: x = 2.
2. 10 mL gives 30 mL steam, so y/2 = 3 and y = 6.
3. The hydrocarbon is \( \ce{C2H6 \)}.

Teacher's Note:
1. At the same temperature and pressure, volume ratios equal mole ratios (Gay-Lussac).
2. Carbon atoms come from \( \ce{CO2 \)}, hydrogen atoms from \( \ce{H2O \)}.

Answer: (b) 4 moles of C

Explanation:
1. 5 mol of A would need 10 mol of B, but only 8 mol of B is available, so B is the limiting reagent.
2. 2 mol of B gives 1 mol of C, so 8 mol of B gives 4 mol of C.
3. Product formed = 4 mol of C.

Teacher's Note:
1. The limiting reagent decides the amount of product.
2. Divide moles of each reactant by its coefficient; the smaller value limits the reaction.

Answer: (b) \( 12.044\times10^{23 \)}

Explanation:
1. The Avogadro number is the number of atoms in 12 g of C-12 on the present scale, i.e. the number of atoms whose total mass equals the atomic mass expressed in grams.
2. If carbon were assigned 24 amu, every atom would be given twice its present relative mass, so the mole (24 g of carbon) would contain twice as many atoms.
3. Avogadro constant = 2 x 6.022 x 10²³ = \( 12.044\times10^{23 \)}.

Teacher's Note:
1. The mole is defined by relative atomic mass scale; changing the standard changes the constant.
2. The question is conceptual; no calculation with actual masses is needed.

Answer: (b) 20

Explanation:
1. Let the abundance of B-10 be x%. Then 10x + 11(100 - x) = 1080.
2. 1100 - x = 1080, so x = 20.
3. Abundance of B-10 = 20%.

Teacher's Note:
1. Average atomic mass is the abundance-weighted mean of isotopic masses.
2. The average (10.80) lies closer to 11, so B-11 must be the more abundant isotope.

Answer: (a) \( \ce{C8H16O4 \)}

Explanation:
1. Oxygen = 100 - 54.54 - 9.09 = 36.37%. Mole ratio C : H : O = 4.545 : 9.09 : 2.273 = 2 : 4 : 1, so the empirical formula is \( \ce{C2H4O \)} (mass 44).
2. Molar mass = 2 x vapour density = 176, so n = 176/44 = 4.
3. Molecular formula = \( \ce{C8H16O4 \)}.

Teacher's Note:
1. Molar mass = 2 x vapour density.
2. Molecular formula = n x empirical formula, n = molar mass / empirical mass.

Answer: (d) \( \ce{X2Y3 \)}

Explanation:
1. Take 100 g: X = 76 g and Y = 24 g.
2. Moles of X = 76/75 = 1.013 and of Y = 24/16 = 1.5; the ratio is 1.013 : 1.5 = 2 : 3.
3. Formula = \( \ce{X2Y3 \)}.

Teacher's Note:
1. Assume 100 g of compound to turn percentages into masses.
2. Divide by atomic masses and reduce to the simplest whole-number ratio.

Answer: (c) 3.52 g

Explanation:
1. Pure \( \ce{CaCO3 \)} = 40% of 20 g = 8 g = 0.08 mol.
2. \( \ce{CaCO3 -> CaO + CO2 \)}: 1 mol gives 1 mol \( \ce{CO2 \)}, so 0.08 mol \( \ce{CO2 \)} forms.
3. Mass = 0.08 x 44 = 3.52 g.

Teacher's Note:
1. Always take only the pure part of an impure sample into the stoichiometry.
2. Limestone is mainly calcium carbonate.

Answer: (b) 49 g

Explanation:
1. \( \ce{CaCO3 + H2SO4 -> CaSO4 + H2O + CO2 \)}. 25 g \( \ce{CaCO3 \)} = 0.25 mol.
2. 0.25 mol of \( \ce{H2SO4 \)} = 0.25 x 98 = 24.5 g of pure acid.
3. The acid is 50% pure, so mass of solution needed = 24.5/0.5 = 49 g.

Teacher's Note:
1. Moles of acid equal moles of carbonate (1:1).
2. Divide by the purity fraction to get the mass of the solution.

Answer: (a) \( \ce{O2 \)} is the limiting reagent

Explanation:
1. 8 g \( \ce{H2 \)} = 4 mol; 32 g \( \ce{O2 \)} = 1 mol. \( \ce{2H2 + O2 -> 2H2O \)}.
2. 1 mol of \( \ce{O2 \)} needs only 2 mol of \( \ce{H2 \)}, so \( \ce{H2 \)} is in excess and \( \ce{O2 \)} is the limiting reagent.
3. Water formed = 2 mol = 36 g, so 40 g is wrong. Correct statement: \( \ce{O2 \)} is limiting.

Teacher's Note:
1. Convert masses to moles, then compare with the equation's coefficients.
2. Water formed is fixed by the limiting reagent.

Answer: (a) 941.176

Explanation:
1. Minimum molecular weight corresponds to one sulphur atom per molecule.
2. Molar mass = 32/0.034 = 941.18.
3. Minimum molecular weight = 941.176.

Teacher's Note:
1. With more sulphur atoms the molar mass would be a whole-number multiple, so one atom gives the minimum.
2. Mass of S / fraction of S = molar mass.

Answer: (a) 15 L \( \ce{H2 \)} gas at STP

Explanation:
1. Convert each to moles: 15/22.4 = 0.67 mol; 5/22.4 = 0.22 mol; 0.5/2 = 0.25 mol; 10/32 = 0.31 mol.
2. The largest is 0.67 mol, for 15 L of \( \ce{H2 \)}.
3. 15 L \( \ce{H2 \)} at STP has the most molecules.

Teacher's Note:
1. Number of molecules is proportional to moles.
2. Use 22.4 L/mol for gases at STP and molar mass for masses.

Answer: (c) \( \dfrac{y}{x-y}\times35.5 \)

Explanation:
1. Mass of chlorine = (x - y) g.
2. Equivalent weight of metal combines with 35.5 g of chlorine: y/(x - y) = E/35.5.
3. E = \( \dfrac{y}{x-y}\times35.5 \).

Teacher's Note:
1. Equivalent weight is the mass of the element that combines with 35.5 g of chlorine.
2. Check by taking a simple case such as NaCl.

Answer: (a) 56 : 4 : 7

Explanation:
1. For equal mass w, moles = w/M, so the volumes are in the ratio 1/2 : 1/28 : 1/16.
2. Multiplying by 112 gives 56 : 4 : 7.
3. Ratio = 56 : 4 : 7.

Teacher's Note:
1. At the same T and P, volume is proportional to moles.
2. The lightest gas occupies the largest volume for equal mass.

Answer: (d) 1.12 L, 1.12 L

Explanation:
1. \( \ce{H2 + Cl2 -> 2HCl \)}. 3.65 g HCl = 0.1 mol.
2. 0.1 mol HCl needs 0.05 mol each of \( \ce{H2 \)} and \( \ce{Cl2 \)}.
3. Volume of each = 0.05 x 22.4 = 1.12 L.

Teacher's Note:
1. Gases react in the ratio of their coefficients.
2. 1 mol of gas at NTP = 22.4 L.

Answer: (a) 1 g-atom of C

Explanation:
1. 1 g-atom of C = 12 g.
2. 0.5 mol \( \ce{CH4 \)} = 8 g; 10 mL water = 10 g; 3.011 x 10²³ O atoms = 0.5 mol x 16 = 8 g.
3. The highest mass is 12 g (1 g-atom of C).

Teacher's Note:
1. Convert every option into grams before comparing.
2. Density of water is 1 g/mL.

Answer: (c) 2 litre

Explanation:
1. \( \ce{2CO + O2 -> 2CO2 \)}.
2. Volume ratio CO : \( \ce{O2 \)} = 2 : 1.
3. \( \ce{O2 \)} needed = 4/2 = 2 litre.

Teacher's Note:
1. For gases at the same conditions, coefficients give volume ratios directly.
2. No need to convert to moles.

Answer: (c) 5

Explanation:
1. Molar mass = 2 x 70 = 140 g/mol.
2. Mass of one CO unit = 28, so x = 140/28 = 5.
3. x = 5.

Teacher's Note:
1. Vapour density is half the molar mass.
2. The molecular formula is a whole-number multiple of the empirical unit.

Answer: (a) \( \ce{C3H8 \)}

Explanation:
1. Equal volumes of gases contain equal moles (Avogadro's law), so equal weights mean equal molar masses.
2. Molar mass of \( \ce{CO2 \)} = 44 g/mol.
3. Only \( \ce{C3H8 \)} (36 + 8 = 44) matches, so the hydrocarbon is \( \ce{C3H8 \)}.

Teacher's Note:
1. Equal volume and equal mass mean equal molar mass.
2. Check each option by adding atomic masses.

Answer: (c) 2 : 1 : 2 : 3

Explanation:
1. Equal volumes contain equal numbers of molecules, say N.
2. Atoms per molecule: \( \ce{H2 \)} = 2, He = 1, \( \ce{O2 \)} = 2, \( \ce{O3 \)} = 3.
3. Ratio of atoms = 2 : 1 : 2 : 3.

Teacher's Note:
1. Helium is monoatomic, ozone is triatomic.
2. Use Avogadro's law first, atomicity second.

Answer: (d) 3 : 2

Explanation:
1. First oxide: 50 g S per 50 g O, so 1 g S per 1 g O.
2. Second oxide: 40 g S per 60 g O, so 2/3 g S per 1 g O.
3. Ratio = 1 : 2/3 = 3 : 2.

Teacher's Note:
1. This illustrates the law of multiple proportions.
2. Fix the weight of one element (oxygen) and compare the other.

Answer: (d) \( \ce{NO2 \)}

Explanation:
1. Moles N = 2.34/14 = 0.167; moles O = 5.34/16 = 0.334.
2. Ratio N : O = 1 : 2.
3. Simplest formula = \( \ce{NO2 \)}.

Teacher's Note:
1. Divide each mass by its atomic mass, then by the smallest value.
2. Simplest formula is the whole-number mole ratio.

Answer: (b) 32 cc

Explanation:
1. \( \ce{CO2 + C -> 2CO \)}.
2. 1 volume of \( \ce{CO2 \)} gives 2 volumes of CO.
3. Volume of CO = 2 x 16 = 32 cc.

Teacher's Note:
1. Solid carbon does not occupy gas volume.
2. Gas volumes follow the coefficients.