Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
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$T_1\,+\, T_2$
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$(T_1\, +\, T_2)/2$
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$\displaystyle \frac{T_1\, T_2\, P(V_1\, +\, V_2)}{P_1V_1T_2\, +\, P_2V_2T_1}$
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$\displaystyle \frac{T_1\, T_2\, (P_1V_1\, +\, P_2V_2)}{P_1V_1T_2\, +\, P_2V_2T_1}$
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$0.22kg/m^{3}$
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$0.62kg/m^{3}$
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$1.12kg/m^{3}$
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$0.13kg/m^{3}$
D
Correct answer
Explanation
At NPT, 1 mole of any gas occupies 22.4 L. Hydrogen (H2): 4g = 2 moles. Helium (He): 8g = 2 moles. Total moles = 4. Total volume = 4 * 22.4 L = 89.6 L = 0.0896 m^3. Total mass = 4g + 8g = 12g = 0.012 kg. Density = 0.012 / 0.0896 = 0.1339 kg/m^3.
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4gm,24gm
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1gm,27gm
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6gm,22gm
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2gm,26gm
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$45.4$
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$49.8$
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$32.6$
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$38.3$
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$18mL$
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$12mL$
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$23mL$
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$121mL$
B
Correct answer
Explanation
Henry's Law: Solubility is proportional to partial pressure. Initial: 10 mL/L at 1 atm. New partial pressure of H2 = (68.5/100) * (1400/760) atm = 0.685 * 1.842 = 1.26 atm. New solubility = 10 * 1.26 = 12.6 mL/L. 12 mL is the closest option.
A
Correct answer
Explanation
Vapour density = M_mix / 2. M_mix = 34.5 * 2 = 69. Let x be the mole fraction of NO2 (M=46) and (1-x) be the mole fraction of N2O4 (M=92). 46x + 92(1-x) = 69. 46x + 92 - 92x = 69. -46x = -23. x = 0.5. So 50%.
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$88\ g$
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$67\ g$
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$54\ g$
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$45\ g$
B
Correct answer
Explanation
Heat lost by water cooling to 0C: 200 * 4.2 * 15 = 12600 J. Heat gained by ice warming to 0C: 100 * 2.2 * 7 = 1540 J. Remaining heat available to melt ice: 12600 - 1540 = 11060 J. Mass of ice melted: 11060 / 335 = 33.01 g. Remaining ice: 100 - 33.01 = 66.99 g, which is approximately 67 g.
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$20\%$
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$40\%$
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$60\%$
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$45.45\%$
B
Correct answer
Explanation
Let E be mass of ethanol, W be mass of water. E + W = 50. E/0.8 + W/1.0 = 55. Substitute W = 50 - E: E/0.8 + (50-E) = 55. 1.25E - E = 5, 0.25E = 5, E = 20. Percentage = 20/50 = 40%.
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$21.12$%
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$78.88$%
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$5$%
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$95$%
B
Correct answer
Explanation
Let x be the mass of AgCl and y be the mass of AgBr. The molar masses are approximately AgCl=143.3 and AgBr=187.8. Chlorination converts AgBr to AgCl, resulting in a mass loss of (187.8 - 143.3) = 44.5 per mole of AgBr. The 5% loss implies 0.05(x + y) = (44.5/187.8)y. Solving this ratio leads to the percentage of AgCl.
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$85\%$
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$72\%$
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$65\%$
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$90\%$
B
Correct answer
Explanation
This is a chemistry stoichiometry problem. 2NH3 -> N2 + 3H2. 50ml mixture. Let NH3 be x, H2 be 50-x. N2 produced = x/2, H2 produced = 3x/2. Total H2 = 50-x + 3x/2 = 50 + x/2. O2 added = 40ml. 2H2 + O2 -> 2H2O. H2 reacts with O2 in 2:1 ratio. 40ml O2 reacts with 80ml H2. Contraction = 6ml. This requires detailed calculation of gas volumes.
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$41.6\%$
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$58.4\%$
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$41.2\%$
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$50\%$
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$82.38$
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$26.38$
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$62.38$
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$28.38$
B
Correct answer
Explanation
Using Charles's Law at constant pressure, V1/T1 = V2/T2. Temperatures must be in Kelvin: T1 = 27+273 = 300K, T2 = 477+273 = 750K. V2 = V1 * (750/300) = 2.5 * V1. The volume of air expelled is V2 - V1 = 1.5 * V1. The fraction expelled is (1.5 * V1) / (2.5 * V1) = 1.5/2.5 = 3/5.
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$40\% A, 60\% B$
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$60\% A, 40\% B$
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$30\% A, 70\% B$
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$70\% A, 30\% B$
A
Correct answer
Explanation
560 mL of H2 at STP is 0.025 moles, which corresponds to 0.05 equivalents. Let mass of A be x, then mass of B is 0.5-x. Equivalents equation: x/12 + (0.5-x)/9 = 0.05. Solving gives x = 0.2, so A is 0.2/0.5 = 40% and B is 60%.