JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 37

Answer: (c) \( 10^{-3}N_A \)

Explanation:
1. The conditions are STP, where 1 mol of gas occupies 22400 cc.
2. Moles present = \( \frac{22.4}{22400}=10^{-3} \).
3. Number of molecules = \( 10^{-3}N_A \).

Teacher's Note:
1. At STP 1 mol of any gas occupies 22.4 L = 22400 cc.
2. Convert the volume to the same unit before dividing.

Answer: (a) 50 mL increase

Explanation:
1. 1 volume of PH3 gives 1.5 volumes of H2 (phosphorus is a solid and occupies no gas volume).
2. 100 mL of PH3 gives \( 100\times1.5=150 \) mL of H2.
3. Change in volume = \( 150-100=+50 \) mL, an increase.

Teacher's Note:
1. Solids do not count in gas-volume changes.
2. Gay-Lussac's law lets us treat coefficients as volume ratios for gases at the same T and P.

Answer: (d) \( 3\times10^{-26} \) kg

Explanation:
1. One mole of water is 18 g and contains \( 6.022\times10^{23} \) molecules.
2. Mass of one molecule = \( \frac{18}{6.022\times10^{23}}=2.99\times10^{-23} \) g.
3. In kg this is \( 2.99\times10^{-26}\approx3\times10^{-26} \) kg.

Teacher's Note:
1. Remember the unit conversion: 1 g = \( 10^{-3} \) kg.
2. Option (a) has the right digits but a wrong power of ten, a common slip.

Answer: (b) \( 6.023\times10^{25} \)

Explanation:
1. Moles of Au = \( \frac{19700\ \text{g}}{197\ \text{g mol}^{-1}}=100 \) mol.
2. Atoms = \( 100\times6.023\times10^{23}=6.023\times10^{25} \).

Teacher's Note:
1. Convert kg to g first.
2. The answer 100 is the number of moles, not atoms.

Answer: (b) 8.4 g

Explanation:
1. Molar mass of \( \ce{MnO2} \) = \( 55+32=87 \) g mol\(^{-1}\); moles = \( 5.0/87=0.0575 \).
2. 4 mol of HCl react per mole of MnO2: \( 4\times0.0575=0.230 \) mol HCl.
3. Mass of HCl = \( 0.230\times36.5=8.4 \) g.

Teacher's Note:
1. Use the coefficients of the balanced equation to link moles of MnO2 to moles of HCl.
2. Do not forget the factor 4 for HCl.

Answer: (d) \( 6.02\times10^{18} \)

Explanation:
1. Mass of S in 1 g = \( \frac{0.032}{100}=3.2\times10^{-4} \) g.
2. Moles of S = \( \frac{3.2\times10^{-4}}{32}=10^{-5} \) mol.
3. One S atom per molecule, so molecules = \( 10^{-5}\times6.02\times10^{23}=6.02\times10^{18} \).

Teacher's Note:
1. The number of S atoms equals the number of molecules here.
2. Take care with the percentage: 0.032% means 0.032 g per 100 g.

Answer: (a) 165.5 g

Explanation:
1. Moles: PbO = \( 446/223=2 \); NO2 = \( 46/46=1 \); O2 = \( 16/32=0.5 \).
2. NO2 needs 2 mol per mol of product, so it is the limiting reagent: product = \( 1/2=0.5 \) mol (PbO and O2 are in excess).
3. Molar mass of Pb(NO3)2 = 331 g mol\(^{-1}\), so mass = \( 0.5\times331=165.5 \) g.

Teacher's Note:
1. Always identify the limiting reagent by dividing each mole amount by its coefficient.
2. The value 331 g would be the answer only if PbO were limiting with 1 mol.

Answer: (c) \( \ce{C2H4O} \)

Explanation:
1. Carbon: 88 g CO2 = 2 mol, so 24 g C. Hydrogen: 36 g H2O = 2 mol, so 4 g H.
2. Oxygen = \( 44-24-4=16 \) g = 1 mol.
3. Mole ratio C : H : O = 2 : 4 : 1, empirical formula C2H4O with mass 44, equal to the given mass of 1 mol, so the formula is \( \ce{C2H4O} \).

Teacher's Note:
1. Mass balance gives oxygen by difference.
2. Check the formula mass against the sample mass for 1 mol.

Answer: (c) All of the above

Explanation:
1. 0.22 g CO2 = \( 0.22/44=0.005 \) mol.
2. 0.01 g H2 = 0.005 mol; 0.085 g NH3 = \( 0.085/17=0.005 \) mol; 320 mg SO2 = \( 0.32/64=0.005 \) mol.
3. Equal moles occupy equal volumes at the same T and P, so all three match.

Teacher's Note:
1. Equal volumes at the same T and P mean equal moles (Avogadro's law).
2. Convert mg to g before dividing by molar mass.

Answer: (a) 1120 cc of carbon dioxide at STP

Explanation:
1. 0.1 g atom of C = \( 0.1\times12=1.2 \) g.
2. 0.1 mol NH3 = \( 0.1\times17=1.7 \) g.
3. \( 6.02\times10^{22} \) molecules of H2 = 0.1 mol = 0.2 g.
4. 1120 cc CO2 = 0.05 mol = \( 0.05\times44=2.2 \) g, the largest.

Teacher's Note:
1. Convert every quantity to grams to compare.
2. CO2 has the highest molar mass here, so even a smaller number of moles wins.

Answer: (d) 49.5 g

Explanation:
1. The reaction is \( \ce{NaCl + H2SO4 -> NaHSO4 + HCl} \).
2. Mass of products = \( 120+27.5=147.5 \) g.
3. Mass of NaCl = \( 147.5-98=49.5 \) g.

Teacher's Note:
1. Total mass of reactants equals total mass of products.
2. No mole calculation is needed here, only the conservation of mass.

Answer: (b) 1 : 16 : 2

Explanation:
1. Take 1 g of each. Moles of O2 = \( 1/32 \), H2 = \( 1/2 \), CH4 = \( 1/16 \).
2. Multiply by 32: 1 : 16 : 2.

Teacher's Note:
1. For equal masses, moles are inversely proportional to the molar masses 32 : 2 : 16.
2. Clear fractions by multiplying by the largest molar mass.

Answer: (b) 20 cc

Explanation:
1. NH3 and HCl react in 1 : 1 volume ratio to form solid NH4Cl.
2. 20 cc NH3 consumes 20 cc HCl, so 20 cc of HCl remains.
3. NH4Cl is a solid and adds nothing to the gas volume.

Teacher's Note:
1. Find the limiting gas first; the excess gas is what remains.
2. Solid products do not count in the final gas volume.

Answer: (d) 55.8

Explanation:
1. \( \ce{X2O3 + 3H2 -> 2X + 3H2O} \). 6 mg H2 = 0.003 mol.
2. Moles of oxide = \( 0.003/3=0.001 \) mol, so molar mass = \( 0.1596/0.001=159.6 \) g mol\(^{-1}\).
3. \( 2x+48=159.6 \), hence \( x=55.8 \).

Teacher's Note:
1. Three moles of H2 are needed per mole of oxide.
2. Subtract the oxygen contribution (3 x 16) before dividing by 2.

Answer: (a) 1000

Explanation:
1. Atomic mass = \( 6.664\times10^{-23}\times6.022\times10^{23}\approx40 \) g mol\(^{-1}\).
2. Gram atoms = \( \frac{40000\ \text{g}}{40\ \text{g}}=1000 \).

Teacher's Note:
1. One gram atom is one mole of atoms, the atomic mass in grams.
2. Convert 40 kg to 40000 g.

Answer: (c) 0.24

Explanation:
1. Moles of product possible: from A \( 1/4=0.25 \); from B \( 0.6/2=0.30 \); from C \( 0.72/3=0.24 \).
2. C gives the smallest value, so C is the limiting reagent and 0.24 mol of product forms.

Teacher's Note:
1. Divide each amount by its coefficient; the smallest quotient identifies the limiting reagent.

Answer: (c) 4/3

Explanation:
1. Take 2 P atoms in each: that is 1 formula unit of Ca3(PO4)2 and 2 of H3PO3.
2. Oxygen atoms: Ca3(PO4)2 has 8; two H3PO3 have \( 2\times3=6 \).
3. Ratio = \( 8/6=4/3 \).

Teacher's Note:
1. Equalise the number of P atoms first, then count O.
2. Write each formula unit carefully: phosphate has 4 O per P, phosphite only 3.

Answer: (a) \( \frac{X}{2},\ \frac{Y}{3},\ \frac{Z}{6} \)

Explanation:
1. Equivalent weight of a salt = molar mass divided by the total positive (or negative) charge per formula unit.
2. Na2SO4: charge 2 (2 Na\(^+\)), so X/2.
3. Na3PO4.12H2O: 3 Na\(^+\), so Y/3.
4. Ca3(PO4)2: 3 Ca\(^{2+}\) = total charge 6, so Z/6.

Teacher's Note:
1. The n-factor of a salt equals the total charge on the cation (or anion) in one formula unit.
2. Water of crystallisation does not change the n-factor.

Answer: (d) 0.4375 kg

Explanation:
1. \( \ce{CaC2 -> C2H2 -> C2H4 -> (C2H4)n} \): 1 mol CaC2 gives 1 mol of C2H4 units.
2. Moles of CaC2 in 1 kg = \( 1000/64=15.625 \).
3. Mass of polythene = \( 15.625\times28=437.5 \) g = 0.4375 kg.

Teacher's Note:
1. Follow the carbon chain: every step keeps a 1 : 1 mole ratio.
2. The polymer mass equals the total mass of monomer units.

Answer: (b) \( \ce{C2H4O} \)

Explanation:
1. Let %H = h; then %C = 6h and %O = \( 6h/1.5=4h \). Total \( 11h=100 \), so h = 9.09.
2. %C = 54.5, %H = 9.09, %O = 36.4.
3. Moles: C \( 54.5/12=4.54 \), H 9.09, O \( 36.4/16=2.27 \). Divide by 2.27: C : H : O = 2 : 4 : 1.

Teacher's Note:
1. Express all percentages in terms of one variable.
2. Divide by the smallest mole value to get the empirical formula.

Answer: (b) 4 : 1

Explanation:
1. Let the mixture have x mol HCOOH and y mol H2C2O4 (H2O is a liquid).
2. Gas formed = x (CO) + 2y (CO + CO2); KOH absorbs only CO2 = y.
3. \( \frac{y}{x+2y}=\frac{1}{6} \), so \( 6y=x+2y \) and \( x=4y \). Ratio HCOOH : H2C2O4 = 4 : 1.

Teacher's Note:
1. KOH absorbs CO2 but not CO.
2. Liquid water adds nothing to the gas volume.

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

Explanation:
1. \( \ce{C_nH_{2n+2} + \frac{3n+1}{2}O2 -> nCO2 + (n+1)H2O} \).
2. \( \frac{(3n+1)/2}{n}=\frac{7}{4} \), so \( 6n+2=7n \) and \( n=2 \).
3. The alkane is \( \ce{C2H6} \).

Teacher's Note:
1. For an alkane, oxygen required = \( (3n+1)/2 \) mol per mole.
2. Check: 3.5 mol O2 to 2 mol CO2 = 7 : 4.

Answer: (a) \( \ce{M2O3} \)

Explanation:
1. Oxide 1: 77.78 g M with 22.22 g O in the ratio 1 : 1 by moles, so \( M=\frac{77.78}{22.22}\times16=56 \).
2. Oxide 2: 70 g M = \( 70/56=1.25 \) mol and 30 g O = \( 30/16=1.875 \) mol.
3. M : O = 1.25 : 1.875 = 2 : 3, so the formula is \( \ce{M2O3} \).

Teacher's Note:
1. Use the first oxide (MO) to find the atomic mass of the metal.
2. This illustrates the law of multiple proportions.