What quantity (in mL) of a 45% acid solution of a monoprotic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?
Option: 1 320
Option: 2 325
Option: 3 316
Option: 4 330
 

Answers (1)

 

Reactions in Solutions -

Concentration :
It is the amount of solute present in one litre of solution. It is denoted by C or S.

\\ \mathrm{C\, or\, S\, =\, \frac{Weight\, of\, solute\, in\, gram}{Volume\, in\, litre}} \\\\ \mathrm{C\, =\, N\, \times\, E}\\\\\mathrm{Here\, N\, =\, normality\, and \, E\, =\, Eq.\, wt.}

Mole Fraction:
It is the ratio of moles of one component to the total number of moles present In the solution. It is expressed by X for example, for a binary solution with two components A and B.

X_{A}\, = \frac{n_{A}}{n_{A}\, +\, n_{B}}

X_{B}\, = \frac{n_{B}}{n_{A}\, +\, n_{B}}

X_{A}\, +\, X_{B}\, =\, 1

Here nA and nB represent moles of solvent and solute respectively. Mole fraction does not depend upon temperature as both solute and solvent are expressed by weight.

Molarity:
It is the number of moles or gram moles of solute dissolved per litre of the solution. Molarity is denoted by 'M'.
 
\mathrm{M\, =\, \frac{Weight\, of\, solute\, in\, gram}{Molar\, mass\, \times\, volume\, in\, litre}}

  • When molarity of a solution is one, it is called a molar solution and when it is 0.1, solution is called decimolar solution. 
  •  Molarity depends upon temperature and its unit is mol/litre. 
    M1V1 = M2V2
  • On dilution water added = V2 - V1
  • When the solution of two different substances react together then 
    \mathrm{\frac{M_{1}V_{1}}{n_{1}}\, =\, \frac{M_{2}V_{2}}{n_{2}}}
    Here M, V, n are molarity, volume and number of molecules taking part in a reaction respectively. 
  • When a mixture of different solutions having different concentrations are taken the molarity of the mixture is calculated as follows: 
    \mathrm{M\, =\, \frac{M_{1}V_{1}\, + M_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}
  • When density and % by weight of a substance in a solution are given, molarity is find as follows: 
    \mathrm{M\, =\,\frac{\%\, by\, weight\,\times\, d\, \times\, 10}{Molecular\, weight}}
    Here d = density 

Molality
It is the number of moles or gram moles of solute dissolved per kilogram of the solvent. It is denoted by 'm'.
\mathrm{m\, =\, \frac{Weight\, of\, solute\, in\, gram }{Molar\, mass\, \times\, wt.\, of\, solvent\, in\, Kg}}

  • If molality is one solution, it is called molal solution.
  • One molal solution is less than one molar solution.
  • Molality is preferred over molarity during experiments as molality is temperature independent while molarity is temperature-dependent. 

Normality
It is the number of gram equivalents of solute present in one litre of the solution and it is denoted by 'N'.
\mathrm{N\, =\, \frac{Weight\, of\, solute\, in\, gram }{Equivalent\, mass\, \times\, volume\, in\, litre }}

  • When normality of a solution is one, the solution is called normal solution and when it is 0.1, the solution is called deci-normal solution. 

Normality Equation:

\mathrm{N_{1}V_{1}\, =\, N_{2}V_{2}}

  • Volume Of water added = V2 - V1 
    Here V2 = volume after dilution 
            V1 = volume before dilution 
  • When density and % by weight of a substance in a solution are given, normality is find as follows: 

    \mathrm{N\, =\, \frac{\%\, by\, weight\, \times\, d \times\, 10 }{Equivalent\ weight }}
    Here d = density of solution 
  • When a mixture of different solutions having different concentrations are taken the normality of the mixture is calculated as follows: 
    \mathrm{N\, =\, \frac{N_{1}V_{1}\, + N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}
    • In case of acid-base neutralization the normality of the resulting solution 
      \mathrm{N\, =\, \frac{N_{1}V_{1}\, - \, N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}
    • To find weight of substance 
    W\, =\, \frac{NEV}{1000}

Relation between Normality and Molarity :
N x Eq wt. = molarity x molar mass 
N = molarity x valency 
N = molarity x number of H+ or OH- ion 

-

 

 

As we have learnt,

\\\mathrm{\frac{V\times45}{100}\,+\,\frac{(800-V)20}{100}\, =\, \frac{800\times29.875}{100}}\\\\\mathrm{\frac{9V}{20}\, +\, 160\, -\, \frac{V}{5}\, =\, 239}\\\\\mathrm{\frac{5V}{20}\, =\, 79\,}\\\\\mathrm{\therefore V\, =\, 316}

Therefore, Option (3) is correct

 

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