Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
D
Correct answer
Explanation
Partial pressure is proportional to the number of moles. Moles: O2 = 8/32 = 0.25, N2 = 7/28 = 0.25, H2 = 1/2 = 0.5. Total moles = 1.0. Total pressure = 8 atm. Pressure per mole = 8 atm. Removing O2 (0.25 moles) leaves 0.75 moles. Remaining pressure = 0.75 * 8 = 6 atm.
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$\dfrac{1}{3}$ of total pressure
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$\dfrac{2}{3}$ of total pressure
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$\dfrac{3}{2}$ of total pressure
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$\dfrac{1}{2}$ of total pressure
B
Correct answer
Explanation
Moles of H2 = 3/2 = 1.5. Moles of O2 = 24/32 = 0.75. Total moles = 2.25. Mole fraction of H2 = 1.5 / 2.25 = 2/3. Partial pressure = mole fraction * total pressure.
D
Correct answer
Explanation
Molar mass of CO = 28, N2 = 28. Since both have the same molar mass, the mixture's average molar mass is 28. Vapour density = Molar mass / 2 = 28 / 2 = 14.
A
Correct answer
Explanation
Moles of H2 = 2/2 = 1. Moles of O2 = 8/32 = 0.25. Total moles = 1.25. Mole fraction of O2 = 0.25 / 1.25 = 1/5 = 0.2.
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the ratio of their molar masses is $16:1$
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the ratio of their molar masses is $1:4$
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the ratio of their molar present inside the container is $1:24$
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the ratio of their moles present inside the container is $8:3$
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$120$ mL
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$140$ mL
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$144$ mL
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$168$ mL
C
Correct answer
Explanation
Red = 1/4 total. Ratio Red:Blue = 3:2, so Blue = (2/3) * Red = (2/3) * (1/4) = 1/6 total. Yellow = 1 - (1/4 + 1/6) = 1 - 5/12 = 7/12 total. 7/12 * Total = 84, so Total = 144 mL.
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$21.6$ atm
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$0.126$ atm
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$2.16$ atm
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$0.216$ atm
C
Correct answer
Explanation
Osmotic pressure is a colligative property. For a mixture, Pi_final = (Pi1*V1 + Pi2*V2) / (V1 + V2). Pi_final = (1.5*1 + 2.5*2) / (1 + 2) = (1.5 + 5) / 3 = 6.5 / 3 = 2.166 atm.
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0.167M
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0.0167M
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0.125M
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0.0833M
C
Correct answer
Explanation
Using M1V1 = M2V2: 0.25 * 250 = M2 * 500. M2 = (0.25 * 250) / 500 = 0.25 / 2 = 0.125M.
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125 mm
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500 mm
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1000 mm
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250 mm
D
Correct answer
Explanation
Since the vessels have equal volumes and are joined, the total volume doubles. By Boyle's Law (P1V1 + P2V2 = PfVf), where V1=V2=V and Vf=2V: 100*V + 400*V = Pf*(2V). Thus, 500 = 2*Pf, so Pf = 250 mm.
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$15.2 ^o C$
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$30 ^o C$
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$35.2 ^o C$
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$43.7^o C$
B
Correct answer
Explanation
Heat lost = Heat gained. 10g * (50-T) = 20g * (T-20). 500 - 10T = 20T - 400. 30T = 900. T = 30 degrees C.
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600 mL
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900 mL
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1200 mL
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1500 mL
B
Correct answer
Explanation
Degree of dissociation alpha = sqrt(Ka/C). If alpha doubles, C must become C/4. Initial C = 0.2 M. Final C = 0.05 M. Initial volume = 300 mL. C1V1 = C2V2 => 0.2 * 300 = 0.05 * V2. V2 = 1200 mL. Water to be added = 1200 - 300 = 900 mL.
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$4^{0}C$
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$9.33^{0}C$
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$2.67^{0}C$
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$6.67^{0}C$
C
Correct answer
Explanation
Heat lost = Heat gained. m1 * c * (100 - 27) = m2 * c * (27 - t). Ratio m1:m2 = 1:3. 1 * 73 = 3 * (27 - t). 73 = 81 - 3t. 3t = 8. t = 8/3 = 2.67 degrees Celsius.
A
Correct answer
Explanation
Using the dilution formula M1V1 = M2V2, we have 5 * 3 = 1 * V2, so V2 = 15 L. The volume of water to be added is 15 - 3 = 12 L.
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3740 mm
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1120 mm
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2442 mm
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3464 mm
D
Correct answer
Explanation
Using Boyle's Law (P1V1 = P2V2) for each gas in a 20L flask: P_N2 = (760 * 60) / 20 = 2280 mm. P_H2 = (740 * 32) / 20 = 1184 mm. Total pressure = 2280 + 1184 = 3464 mm.