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

$200 ml$ of ${ O }{ 2 }$ gas maintained at $700mm$ pressure and $250 ml$ of ${ N }{ 2 }$ gas maintained at $720mm$ pressure are put together in one-litre flask.If the temperature is kept constant, the final pressure of the mixture in mm is:
  1. $450$
  2. $320$
  3. $632$
  4. $316$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At constant temperature, each gas contributes pressure equal to its original pressure multiplied by its volume divided by the final volume. The final pressure is 700 x 0.2 + 720 x 0.25 = 320 mm.

Multiple choice

A gaseous mixture contains $\displaystyle CO_{2}\left ( g \right )$ and $\displaystyle N_{2}O\left ( g \right )$ in a $2:5$ ratio by mass. Find the ratio of the number of molecules of $\displaystyle CO_{2}\left ( g \right )$ and $\displaystyle N_{2}O\left ( g \right )$.
  1. $2 : 5$
  2. $5 : 2$
  3. $3 : 5$
  4. $5 : 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass ratio CO2 : N2O = 2 : 5. Molar mass CO2 = 44, N2O = 44. Ratio of moles = (2/44) : (5/44) = 2 : 5. Since number of molecules is proportional to moles, the ratio is 2 : 5.

Multiple choice

A volume of $25 ml$ of $ (N/10)-Na_{2}CO_{3}$ solution neutralizes 10 ml of a dilute $ H_{2}SO_{4}$ solution. The volume of water that must be added to $400 ml$ of this $ H_{2}SO_{4}$ solution in order to make it exactly $N/10$ is
  1. $1000 ml$
  2. $600 ml$
  3. $500 ml$
  4. $400 ml$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

25 ml of N/10 Na2CO3 neutralizes 10 ml of H2SO4. The H2SO4 normality is (25 * 0.1) / 10 = 0.25 N. To dilute 400 ml of 0.25 N to 0.1 N, use N1V1 = N2V2: 0.25 * 400 = 0.1 * V2, so V2 = 1000 ml. Water to add = 1000 - 400 = 600 ml.

Multiple choice

If equal volumes of $0.1M$ $HBr$ and $0.1M$ $KOH$ are mixed then which of the following is/are correct about the resulting solution? (1) $\left[ { H }_{ 3 }{ O }^{ + } \right] =1.0{ \times 10 }^{ -7 }mol\quad { L }^{ -1 }$ (2) $\left[ { OH }^{ - } \right] ={ 1.0\times 10 }^{ -7 }mol\quad { L }^{ -1 }$ (3) $\left[ { K }^{ + } \right] =0.05mol\quad { L }^{ -1 }$ (4) $\left[ { Br }^{ - } \right] =1.0mol\quad { L }^{ -1 }$

  1. 1, 2 and 3 are correct

  2. 1 and 2 are correct

  3. 2 and 4 are correct

  4. 1 and 3 are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mixing equal volumes of 0.1M HBr and 0.1M KOH results in a complete neutralization reaction, forming 0.05M KBr in the final solution. The concentrations of H3O+ and OH- are both 10^-7 M due to the auto-ionization of water, and [K+] is 0.05 M. Statement 4 is incorrect because [Br-] is 0.05 M, not 1.0 M.