Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
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Will remain unchanged in $A$ and $B$
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Will increase in $A$ and decrease in $B$
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Will decrease in $A$ and increase in $B$
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Will increase in both $A$ and $B$
A
Correct answer
Explanation
According to Dalton's Law or simply the nature of ideal gases, if the compartments are equal and the pressure is the same, removing the wall does not change the pressure because the total volume and total moles remain constant relative to the initial state.
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$0.6\ atm$
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$1.2\ atm$
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$2.4\ atm$
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$3.6\ atm$
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$0.007$ atm
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$1.04$ atm
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$0.13$ atm
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$1.97$ atm
B
Correct answer
Explanation
Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of its individual components. Total = 0.03 + 0.66 + 0.35 = 1.04 atm.
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$\dfrac { RT }{ 50 }$
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$\dfrac { RT }{ 100 }$
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$\dfrac { RT }{ 10 }$
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$\dfrac { RT }{ 40 }$
B
Correct answer
Explanation
Gas A contributes 1.6/32 = 0.05 mol, and gas B contributes 2.2/44 = 0.05 mol, giving 0.10 mol in total. From PV = nRT with V = 10 dm^3, P = 0.10RT/10 = RT/100.
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$111\space mm$
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$222\space mm$
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$333\space mm$
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$444\space mm$
D
Correct answer
Explanation
Using Boyle's Law P1V1 = P2V2, the final pressure P = (P1V1 + P2V2) / V_total. P = (200 * 720 + 400 * 750) / 1000 = (144000 + 300000) / 1000 = 444000 / 1000 = 444 mm.
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Half of $740\space mm$
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Unchanged
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$\displaystyle\frac{1}{9}th$ of $740\space mm$
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Double than $740\space mm$
A
Correct answer
Explanation
According to Dalton's Law of Partial Pressures, the total pressure is the sum of partial pressures. If the number of molecules is equal, the partial pressures are equal (740/2 = 370 mm each). Removing oxygen leaves only hydrogen, so the pressure becomes 370 mm, which is half of 740 mm.
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$60$ $mm \:Hg$
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$80$ $mm \:Hg$
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$100$$mm \:Hg$
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$150$ $mm \:Hg$
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$200$ $mm \:Hg$
A
Correct answer
Explanation
Neon initially occupies 300 mL at 100 mm Hg. After expansion into the combined 500 mL volume, its partial pressure is 100 × 300/500 = 60 mm Hg. The oxygen does not affect this calculation.
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$22.4\ litre$
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$11.2\ litre$
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$44.8\ litre$
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$56\ litre$
D
Correct answer
Explanation
Moles of O2 = 32/32 = 1 mol. Moles of H2 = 3/2 = 1.5 mol. Total moles = 2.5 mol. At STP (760 mm, 0C), 1 mole occupies 22.4 liters. 2.5 moles occupy 2.5 * 22.4 = 56 liters.
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$0^\circ C$
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$10^\circ C$
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$-30^\circ C$
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$>10^\circ C$
A
Correct answer
Explanation
Heat lost by water (5g, 30C to 0C) = 5 * 1 * 30 = 150 cal. Heat gained by ice (5g, -20C to 0C) = 5 * 0.5 * 20 = 50 cal. Remaining heat = 100 cal. This melts 100/80 = 1.25g of ice. Since not all ice melts, the final temperature is 0C.
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840 ml
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420 ml
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630 ml
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None of these
A
Correct answer
Explanation
Total moles of H2O2 = (0.1 * 0.1) + (0.4 * 0.5) = 0.01 + 0.2 = 0.21 moles. Reaction: 2KMnO4 + 5H2O2 + 6H+ -> 2Mn2+ + 5O2 + 8H2O. Moles of KMnO4 = (2/5) * moles of H2O2 = 0.4 * 0.21 = 0.084 moles. Volume of 0.1 M KMnO4 = 0.084 / 0.1 = 0.84 litres = 840 ml.
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$750cc$
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$400cc$
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$800cc$
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$600cc$
C
Correct answer
Explanation
Let initial moles be x (CO2) and y (CO). Total = x+y. Reaction: CO2 + C -> 2CO. If x moles of CO2 react, they produce 2x moles of CO. Net change in moles = +x. Increase = x / (x+y) = 0.20. x = 0.2x + 0.2y. 0.8x = 0.2y. y = 4x. Total = 5x. %CO2 = x/5x = 20%.
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$1:2$
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$1:1$
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$1:16$
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$15:16$
D
Correct answer
Explanation
Equal weights means moles of H2 = w/2 and moles of C2H6 = w/30. Mole fraction of H2 = (w/2) / (w/2 + w/30) = (1/2) / (16/30) = (1/2) * (30/16) = 15/16.