Mixtures and Alligation Questions

Multiple choice
  1. 5ml, 2ml, 3ml

  2. 2ml, 6ml, 4ml

  3. 1ml, 8ml, 9ml

  4. 4ml, 3ml, 5ml

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

CO + 0.5 O2 -> CO2 (contraction 0.5 vol). CH4 + 2 O2 -> CO2 + 2 H2O (contraction 2 vol). N2 is inert. Let x, y, z be volumes of CO, CH4, N2. x+y+z=10. 0.5x + 2y = 6.5. x+y=7 (from KOH absorption of CO2). Solving: x=5, y=2, z=3.

Multiple choice
  1. $10\ ml$
  2. $20\ ml$
  3. $30\ ml$
  4. $25\ ml$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Turpentine absorbs O3. 20 ml reduction means 20 ml of the 100 ml is O3. Heating O3 causes it to convert to O2: 2O3 -> 3O2. The 20 ml of O3 will produce 30 ml of O2. The net increase in volume is 30 - 20 = 10 ml.

Multiple choice
  1. $\displaystyle \:NO= 44ml; N_{2}O= 16ml$
  2. $\displaystyle \:NO= 45ml; N_{2}O= 20ml$
  3. $\displaystyle \:NO= 34ml; N_{2}O= 22ml$
  4. $\displaystyle \:NO= 20ml; N_{2}O= 26ml$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reactions: N2O + H2 -> N2 + H2O; 2NO + 2H2 -> N2 + 2H2O. Let x be vol of N2O and y be vol of NO. x + y = 60. From stoichiometry, x + y/2 = 38. Solving: y/2 = 22, y = 44. Then x = 16.

Multiple choice
  1. 4 : 1: 5

  2. 2 : 3: 5

  3. 1 : 4: 5

  4. 1 : 3: 5

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Combustion of CO produces CO2, while CO2 remains unchanged. The contraction of 40 ml corresponds to the volume of O2 consumed (V_CO/2). The KOH treatment removes all CO2. Solving the stoichiometry leads to the ratio 1:4:5.

Multiple choice
  1. $\displaystyle \:N_{2}O$
  2. $\displaystyle \:NO_{2}$
  3. $\displaystyle \:N_{2}O_{3}$
  4. $\displaystyle \:N_{2}O_{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Reaction: N_xO_y + H_2 -> H_2O + N_2. 10 ml of compound + 30 ml H_2 -> 10 ml N_2 + H_2O. By conservation of atoms: 10*x = 2*10 (N atoms), so x=2. 10*y = 30 (O atoms from H_2O), so y=3. Formula is N_2O_3.

Multiple choice
  1. $2:1$
  2. $1: 2$
  3. $1.5:1.5$
  4. $0.5: 2.5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let x be volume of propane and y be volume of butane. x + y = 3. Combustion: C3H8 + 5O2 -> 3CO2 + 4H2O; C4H10 + 6.5O2 -> 4CO2 + 5H2O. So, 3x + 4y = 10. Solving the system: 3x + 3y = 9 and 3x + 4y = 10 gives y = 1, x = 2. Ratio is 2:1.

Multiple choice
  1. $\dfrac{1}{3}$rd of the total volume
  2. $\dfrac{1}{4}$th of the total volume
  3. $\dfrac{2}{3}$rd of the total volume
  4. $\dfrac{1}{2}$nd of the total volume
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reaction: C2H4 + H2 -> C2H6. 1 mole of C2H4 reacts with 1 mole of H2 to form 1 mole of C2H6. The total moles decrease by 1 for every mole of C2H4. Initial pressure = 600. Final pressure = 400. Decrease = 200. This decrease equals the initial moles of C2H4. Fraction = 200/600 = 1/3.

Multiple choice
  1. $74.4$
  2. $70$
  3. $80$
  4. $90$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let x be the mass of CuSO4.5H2O and y be the mass of MgSO4.7H2O. The molar masses are approximately 250 g/mol for CuSO4.5H2O (anhydrous 160) and 246 g/mol for MgSO4.7H2O (anhydrous 120). We have x + y = 5 and (160/250)x + (120/246)y = 3. Solving this system yields x approximately 3.72 g, which is 74.4% of 5 g.

Multiple choice
  1. $x_{C_{2}H_{6}}=0.3293$, $x_{C_{2}H_{4}}=0.6707$,
  2. $x_{C_{2}H_{6}}=0.6707$, $x_{C_{2}H_{4}}=0.3293$,
  3. $x_{C_{2}H_{6}}=0.5$, $x_{C_{2}H_{4}}=0.5$,
  4. none of these

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total moles n = PV/RT = (1 * 40) / (0.0821 * 400) = 1.218 moles. Let x be moles of C2H4 and y be moles of C2H6. x + y = 1.218. Combustion: C2H4 + 3O2 -> 2CO2 + 2H2O; C2H6 + 3.5O2 -> 2CO2 + 3H2O. O2 moles = 130/32 = 4.0625. 3x + 3.5y = 4.0625. Solving gives x = 0.401, y = 0.817. Mole fractions: x_C2H4 = 0.329, x_C2H6 = 0.671.