Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
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$28\%$
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$45\%$
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$40\%$
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$60\%$
B
Correct answer
Explanation
Pehal's solution initially contains 20 g of sugar and 80 g of water, then she adds 20 g of sugar, giving 40 g sugar in 120 g solution. Ishaan's solution contains 50 g sugar, and evaporating 20 g water leaves 50 g sugar in 80 g solution. The mixture has 90 g sugar in 200 g total solution, so its concentration is 45%.
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$4.6 g$
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$1.6 g$
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$2.3 g$
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$23 g$
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10%
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5%
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26.08%
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None of these
C
Correct answer
Explanation
Vapour density (VD) = M_mix / 2. M_mix = 80. Let alpha be the degree of dissociation of N2O4. M_mix = M_initial / (1 + alpha). 80 = 92 / (1 + alpha), so 1 + alpha = 92/80 = 1.15, alpha = 0.15. The mixture contains 1 mole of N2O4 and 2*alpha moles of NO2. Mole fraction of NO2 = 0.3 / 1.15 = 26.08%.
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10.71% in A
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35% in B
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25% in A
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40% in B
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$66.67$
-
$24.02$
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$72.05$
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$69.62$
B
Correct answer
Explanation
Under the same conditions, 1L of each gas contains the same number of moles (Avogadro's Law). Molar mass of O2 = 32 g/mol, CO2 = 44 g/mol. The ratio of masses is 32:44, which simplifies to 8:11.
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$0.33$ $M$
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$0.39$ $M$
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$0.5777$ $M$
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$0.7777$ $M$
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None of the above
A
Correct answer
Explanation
Total moles = (3 * 0.5) + (9 * 0.2777) = 1.5 + 2.4993 = 3.9993. Total volume = 3 + 9 = 12 L. Concentration = 3.9993 / 12 = 0.333275 M, which is approximately 0.33 M.
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$2.3\ g$
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$0.8\ g$
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$1.15\ g$
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$11.5\ g$
C
Correct answer
Explanation
At STP, 22.4 L of any gas is 1 mole. 1.12 L is 0.05 moles. Mixture is equimolar, so 0.025 moles CH4 and 0.025 moles C2H6. Molar masses: CH4=16, C2H6=30. Mass = 0.025 * 16 + 0.025 * 30 = 0.4 + 0.75 = 1.15 g.
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0.33 M
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0.39 M
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0.5777 M
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0.6777 M
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None of the above
A
Correct answer
Explanation
Total moles = (3 * 0.5) + (9 * 0.2777) = 1.5 + 2.4993 = 3.9993. Total volume = 3 + 9 = 12. Concentration = 3.9993 / 12 = 0.333 M.
D
Correct answer
Explanation
Total mass = 5.0 + 20.0 = 25.0 g. Percent of X = (5.0 / 25.0) * 100 = 20%.
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$0.08\ M$
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$0.2\ M$
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$0.1\ M$
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$0.04\ M$
A
Correct answer
Explanation
Using the dilution formula M1V1 = M2V2, where M1=0.4, V1=2.0, and V2 = 2.0 + 8.0 = 10.0. 0.4 * 2.0 = M2 * 10.0, so M2 = 0.8 / 10 = 0.08 M.
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0.10 M
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0.01 M
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0.001 M
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0.0001 M
A
Correct answer
Explanation
Concentration is an intensive property. Dividing a solution into smaller drops does not change the concentration of the solute in the solvent.
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$31.6$
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$21.3$
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$19.2$
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$17.8$
B
Correct answer
Explanation
Let the metals be M1 and M2. The mixture contains M1CO3 and M2CO3. Using the mass loss on heating (which is CO2), one can determine the molar masses and then the weight percentages of the metals.
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$1.6$ L
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$1.8$ L
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$2.3$ L
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$2.8$ L
A
Correct answer
Explanation
Molarity of 98% H2SO4 = (1000 * 1.8 * 0.98) / 98 = 18 M. Using M1V1 = M2V2: 18 * V1 = 2.4 * 12. V1 = (2.4 * 12) / 18 = 1.6 L.
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$m_{CO}: m_{CO_{2}} = 21:11$
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$m_{CO}: m_{CO_{2}} = 11:21$
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$m_{CO}: m_{CO_{2}} = 22:14$
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$m_{CO}: m_{CO_{2}} = 14:22$