Question

The volume of water required to make $0.2M$solution from $16ml$of $0.5M$solution is:
A. $40ml$
B. $16ml$
C. $50ml$
D. 24ml$

Answer 1

Hint: We know that for the same solution if the molarity and volume for one case is given an either the molarity or volume for second case is given then we can calculate the unknown variable as follows: ${M_1}{V_1} = {M_2}{V_2}$.

Complete step by step answer:
First of all we will read about the concentration terms.
Solute: The compound which is mixed in a given substance, is known as solute.
Solvent: The compound which is given and in which solute is mixed, is known as solvent.
Molar Concentration: It is defined as the amount of substance per unit volume of solution.
Molarity: It is defined as the number of moles in the given volume of the solution. The unit of the molarity is as: $mol/litre$.
Molality: It is defined as the number of moles of solute per kilogram of solvent. The unit of molality is: $mol/kg$.
Normality: It is defined as the gram equivalent weight per unit litre of solution. The unit of normality is: $eq/litre$.
Equivalent mass: It is defined as the mass of one equivalent.
If we are given with molarity and volume ( ${M_1},{V_1}$) in case one of solution and for the second case of the solution if we are given with either molarity (${M_2}$) or volume (${V_2}$) then we can find the other parameter by the known by using the relation: ${M_1}{V_1} = {M_2}{V_2}$.
In the given question, we are given with molarity of water in one solution ${M_1} = 0.5M$and the volume of the solution ${V_1} = 16ml$ and molarity of water in second solution ${M_2} = 0.2M$then volume of water in second solution ${V_2}$.
${M_1}{V_1} = {M_2}{V_2}$. Then the volume of water in second solution ${V_2}$ is as
$
  {V_2} = \dfrac{{{M_1}{V_1}}}{{{M_2}}} \\
  {V_2} = \dfrac{{16 \times 0.5}}{{0.2}} \\
  {V_2} = 40ml \\
 $

So the volume of water in the second case ${V_2}$ is $40ml$. Hence the volume of water required to make $16ml$ solution is $40 - 16 = 24ml$.

So, the correct answer is Option D .

Note:
Equivalent mass: It is defined as the mass of one equivalent i.e. the mass of a given substance which will combine with or displace a fixed quantity of another substance. The mass of an element which combines with or displaces $1gram$ of hydrogen or $8gram$of oxygen or $35.5gram$ of chlorine.