Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
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5ml, 2ml, 3ml
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2ml, 6ml, 4ml
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1ml, 8ml, 9ml
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4ml, 3ml, 5ml
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.
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$10\ ml$
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$20\ ml$
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$30\ ml$
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$25\ ml$
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.
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$\displaystyle \:NO= 44ml; N_{2}O= 16ml$
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$\displaystyle \:NO= 45ml; N_{2}O= 20ml$
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$\displaystyle \:NO= 34ml; N_{2}O= 22ml$
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$\displaystyle \:NO= 20ml; N_{2}O= 26ml$
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.
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4 : 1: 5
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2 : 3: 5
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1 : 4: 5
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1 : 3: 5
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.
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$\displaystyle \:N_{2}O$
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$\displaystyle \:NO_{2}$
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$\displaystyle \:N_{2}O_{3}$
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$\displaystyle \:N_{2}O_{5}$
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.
A
Correct answer
Explanation
Nitrogen mass percent in NH4NO3 (MW=80) is 28/80 = 35%. In (NH4)2HPO4 (MW=132), it is 28/132 = 21.21%. Using the mixture rule: 30.40 - 21.21 = 9.19 and 35 - 30.40 = 4.60. The ratio is 9.19/4.60 which is approximately 2:1.
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$2:1$
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$1: 2$
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$1.5:1.5$
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$0.5: 2.5$
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.
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$\dfrac{1}{3}$rd of the total volume
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$\dfrac{1}{4}$th of the total volume
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$\dfrac{2}{3}$rd of the total volume
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$\dfrac{1}{2}$nd of the total volume
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.
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.
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$1.8$ kg
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$2.7$ kg
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$4.5$ kg
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$3.58$ kg
D
Correct answer
Explanation
Combustion of butane (C4H10) and isobutane (C4H10) follows the same stoichiometry: 2C4H10 + 13O2 -> 8CO2 + 10H2O. 1 kg of C4H10 (MW 58) requires (13/2 * 32) / 58 kg of O2, which is approx 3.58 kg.
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$22:3:7$
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$0.5:3:7$
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$1:3:1$
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$1:3:0.5$
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$x_{C_{2}H_{6}}=0.3293$, $x_{C_{2}H_{4}}=0.6707$,
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$x_{C_{2}H_{6}}=0.6707$, $x_{C_{2}H_{4}}=0.3293$,
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$x_{C_{2}H_{6}}=0.5$, $x_{C_{2}H_{4}}=0.5$,
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none of these
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.
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$1/3$
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$1/2$
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$2/3$
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$\cfrac{1}{3}\times \cfrac{273}{308}$
A
Correct answer
Explanation
Equal masses means moles of CH4 = m/16 and moles of O2 = m/32. Total moles = m(1/16 + 1/32) = 3m/32. Mole fraction of O2 = (m/32) / (3m/32) = 1/3. Pressure fraction equals mole fraction.