Quantitative Aptitude

Mixtures and Alligation

1,816 Questions
All Direct removal and addition 448 Alligation cross rule 270 Ratios, quantities, and percentages 233 Weighted mean and concentration 195 Equal quantity replacement 189 Three or more components 174 Repeated equal-volume replacement 154 Profit-based alligation 153

Mixtures and Alligation Questions

Multiple choice

A solution containing $30:g$ of a non-volatile solute in exactly $90:g$ water has a vapour pressure of $21.85: mm: Hg$ at $25^{\circ}C$. Further 18 gm of water is then added to the solution. The resulting solution has a vapour pressure of 22.18 mmHg at $25^{\circ}C$. Calculate (a) molar mass of the solute, and (b) vapour pressure of water at $25^{\circ}C$.
  1. $\;(a)\:30.6\:g\:mol^{-1},\;(b)\:15.53\:mm\:Hg$
  2. $\;(a)\:61.21\:g\:mol^{-1},\;(b)\:23.99\:mm\:Hg$
  3. $\;(a)\:65.45\:g\:mol^{-1},\;(b)\:28.87\:mm\:Hg$
  4. $\;(a)\:84.87\:g\:mol^{-1},\;(b)\:39.98\:mm\:Hg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Raoult's Law: (P0 - P) / P0 = n2 / (n1 + n2). Let M be molar mass of solute. (P0 - 21.85) / P0 = (30/M) / (90/18 + 30/M) = (30/M) / (5 + 30/M) = 30 / (5M + 30). After adding 18g water: (P0 - 22.18) / P0 = (30/M) / (108/18 + 30/M) = (30/M) / (6 + 30/M) = 30 / (6M + 30). Solving the system for M and P0 yields M = 61.21 and P0 = 23.99.

Multiple choice

A radiator was filled with $10:L$ of water to which $2.5:L$ of methanol (density $\,=\,0.8:g.mL^{-1}$) were added. At $9::00:pm$, the vehicle is parked outdoors where the temperature is $0^{\circ}C$. The temperature is decreasing at a uniform rate of $0.5^{\circ}C/min$. Upto what time will there be no danger to the radiator of the car. $K_f:(water)\,=\,1.86:kg.mol^{-1}:K$. Assume methanol to be non-volatile.
  1. $\;11.67\:min$
  2. $\;23.25\:min$
  3. $\;32.56\:min$
  4. $\;65.1\:min$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

At $80^{\mathrm{o}}\mathrm{C}$, the vapour pressure of pure liquid $\mathrm{A}$' is $520\ mm\ Hg$ and that of pure liquid $\mathrm{B}$' is $1000\ mm\ Hg$. If a mixture solution of $\mathrm{A}$' and $\mathrm{B}$' boils at $80^{\mathrm{o}}\mathrm{C}$ and 1 atm pressure, the amount of $\mathrm{A}$' in the mixture is: [$1\ atm=760\ mm\ Hg$]
  1. $48\ mol\%$
  2. $50\ mol\% $
  3. $52\ mol\%$
  4. $34\ mol\% $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Raoult's Law: P_total = xA*PA + xB*PB. 760 = xA*520 + (1-xA)*1000. 760 = 520xA + 1000 - 1000xA. 480xA = 240. xA = 240/480 = 0.5 = 50%.

Multiple choice

A mixture of two immiscible liquids nitrobenzene and water boiling at $97^{o}C$ has a partial vapor pressure of water 733 mm and that of nitrobenzene 27 mm. The ratio of the weights of water to nitrobenzene in the distillate is:

  1. 4

  2. 5

  3. 3

  4. 7
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of weights of two immiscible liquids in distillate is (P1 * M1) / (P2 * M2). P_water = 733, P_nitro = 27. M_water = 18, M_nitro = 123. Ratio = (733 * 18) / (27 * 123) = 13194 / 3321 = 3.97, which is approximately 4.

Multiple choice

Four one–litre flasks are separately filled with the gases hydrogen, helium, oxygen and ozone at the same room temperature and pressure. The ratio of total number of atoms of these gases present in the different flasks would be:

  1. 1:1:1:1

  2. 1:2:2:3

  3. 2:1:2:3

  4. 3:2:2:1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At STP, 1 liter of any gas contains the same number of molecules (Avogadro's Law). The number of atoms is: H2 (2), He (1), O2 (2), O3 (3). The ratio is 2:1:2:3.

Multiple choice

When $10$ mL of ethyl alcohol (density $=0.7893: g: mL^{-1})$ is mixed with $20$ mL of water (density $0.9971: g: mL^{-1})$ at $25^oC$, the final solution has a density of $0.9571 :g :mL^{-1}$. The percentage change in total volume on mixing is :
  1. $3.1\%$
  2. $2.4\%$
  3. $1\%$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial mass = (10 * 0.7893) + (20 * 0.9971) = 7.893 + 19.942 = 27.835 g. Final volume = Mass / Density = 27.835 / 0.9571 = 29.083 mL. Initial volume = 30 mL. Change = 30 - 29.083 = 0.917 mL. Percentage change = (0.917 / 30) * 100 = 3.056%, which rounds to 3.1%.

Multiple choice

A tank contains $40 l $ of solution containing $2 g$ of substance per liter. Salt water containing $3 g$ of this substance per liter runs in at the rate of $4 l/min$ and the well-stirred mixture runs out at the same rate. Find the amount of substance in the tank after $15 minutes$ (in grams).
  1. $110.075$
  2. $121.075$
  3. $111.075$
  4. $211.075$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a differential equation problem: dA/dt = (rate_in * conc_in) - (rate_out * conc_out). dA/dt = (4 * 3) - (4 * A/40) = 12 - A/10. Solving this with A(0) = 80 gives A(t) = 120 - 40e^(-t/10). At t=15, A(15) = 120 - 40e^(-1.5) = 120 - 40(0.2231) = 120 - 8.924 = 111.076.

Multiple choice

$0.1 m^{3}$ of water at $ 80 ^ {0}C$ is mixed with $ 0.3 m^{3} $ of water at $ 60 ^{0}C$.The final temperature of the mixture is
  1. $65^{0}C$
  2. $70^{0}C$
  3. $60^{0}C$
  4. $75^{0}C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the principle of conservation of energy, the heat lost by the hotter water equals the heat gained by the cooler water. Since the specific heat capacity and density of water are constant, the final temperature is the weighted average: (0.1 * 80 + 0.3 * 60) / (0.1 + 0.3) = (8 + 18) / 0.4 = 26 / 0.4 = 65 degrees Celsius.

Multiple choice

200 mL of $N_2$ at $25^o C$ and 250 mm of Hg are mixed with 350 mL of $O_2$ at $25^o C$ and 300 mm of Hg in a vessel of capacity 300 mL. Calculate the final pressure of mixture at $25^o C$.

  1. 326.0 mm Hg

  2. 516.67 mm of Hg

  3. 445.8 mm Hg

  4. 608.0 mm of Hg
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Boyle's Law P1V1 = P2V2 for each gas: N2: 250*200 = P_N2*300 => P_N2 = 166.67. O2: 300*350 = P_O2*300 => P_O2 = 350. Total pressure = 166.67 + 350 = 516.67 mm Hg.

Multiple choice

Ice at $20 C $is added $to \ 50 g$ of water at $40C$.When the temperature of the mixture reaches $0C$,it is found that $20 g$ of ice is still unmelted. The amount of ice added to the water was close to (specific heat of water $=4.2J/g/^oC)$ Specific heat of Ice $=2.1\,J/g/^oC$ Heat of fusion of water at $0^oC=334j/g$
  1. $50g$
  2. $40g$
  3. $60g$
  4. $100g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Interpreting the ice temperature as -20 C, the water releases 50×4.2×40 = 8400 J on cooling to 0 C. For m grams of ice, 20 g melts, so m×2.1×20 + 20×334 = 8400, giving m = 40 g.