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Molarity And Mole Fraction MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Reactions in Solutions is considered one of the most asked concept.

  • 17 Questions around this concept.

Solve by difficulty

How many grams of a concentrated nitric acid solution should be used to prepare 250mL of 2.0M HNO3? The concentrated acid is 70% HNO3.

6.02 \times 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is:

What is the mole fraction of the solute in a 1.00 m aqueous solution ?

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Concepts Covered - 1

Reactions in Solutions

1. Solution:

The solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

2. Solute and Solvent:

The solution is present in the same physical state as that of the solvent.

In case the species forming a solution are all present in the same physical state then the component which is present in a smaller amount is called the solute and the other present in a larger amount is called the solvent.

3. Concentration:

The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution

4. Types of concentration term:

    (I) Mass fraction or % (w/w)

     The mass percentage of a component of a solution is defined as:
\mathrm{Mass\: \%\: of\: a\: component}=\frac{\text { Mass of the component in the solution }}{\text { Total mass of the solution }} \times 100
For example, if a solution is described by 10% glucose in water by mass, it means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution. Concentration described by mass percentage is commonly used in industrial chemical applications. For example, a commercial bleaching solution contains 3.62 mass percentage of sodium hypochlorite in water.

 

(II) Mole fraction: Commonly used symbol for mole fraction is x and subscript used on the right-hand side of x denotes the component.
It is defined as:
\text { Mole fraction of a component }=\frac{\text { Number of moles of the component }}{\text { Total number of moles of all the components }}

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.

 

(III) 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. 

 

 

(IV)  Mass by volume percentage (w/V): Another unit that is commonly used in medicine and pharmacy is mass by volume percentage. It is the mass of solute dissolved in 100 mL of the solution.

 

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

  • When the molarity of a solution is one, it is called a molar solution and when it is 0.1, the solution is called decimolar solution. 
  • Molarity depends upon temperature and its unit is mol/liter. 
  • Number of moles of solute if the Molarity and the Volume in liters are given, is calculated as

Moles = M \times V

In case the volume is given in ml then the millimoles of solute will be given by the above formula

  • On dilution water is added and the final volume is made to V2 , then the moles of solute will remain constant and hence the following formula can be used (V1 is the volume before dilution, V2 is the volume after dilution)

M1V1 = M2V2

  • 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 found as follows: 
    \mathrm{M\, =\,\frac{\%\, by\, weight\,\times\, d\, \times\, 10}{Molecular\, weight}}
    Here d = density 

 

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

  • When the normality of a solution is one, the solution is called a normal solution and when it is 0.1, the solution is called a decinormal 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 found 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}.....}}

 

(VII) Strength :
It is the amount of solute present in one liter 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.}

 

(VIII) The relation between Normality and Molarity :
N = molarity x n-factor

N x Eq wt. = molarity x molar mass 
 

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