Volume of 0.1M ${H_2}S{O_4}$ is required to neutralise 50ml of 0.2M $NaOH$ solution is:
A.25
B.100
C.75
D.50
Answer
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Hint: A solution is a homogeneous mixture of two or more components. It is important to know the extent and applications of a solution but the composition of a solution is also important to determine the feasibility of a reaction thus, to calculate the composition of a solution we must calculate its concentration. For the given question we must use the formula ${N_1}{V_1} = {N_2}{V_2}$.
Complete step by step answer:
A solution is a homogenous mixture of two or more components where the composition of the components is the same. There is a solvent which is present in more quantity and there is a solute which is present in lesser quantity as compared to the solvent.
As we know that it is very important to understand the extent and feasibility and application of a solution but that is possible when we know the concentration of each of the components present in the solution. The composition of a solution is determined by the concentration of the components of a solution. There are various methods to calculate the concentration of these components either quantitatively or qualitatively like mass percentage, volume percentage, mole fraction, molarity, normality and others.
For the above given question we use the formula ${N_1}{V_1} = {N_2}{V_2}$.
Where, ${N_1}$ is the normality of the unknown acid or base, ${N_2}$ is the normality of the known base or acid, ${V_1}$ is the volume of unknown acid or base and ${V_2}$ is the volume of the known base or acid. The formula helps to determine the unknown volume or normality for an acid or base with respect to known normality or volume of an acid or base.
On using the formula ${N_1}{V_1} = {N_2}{V_2}$ for the above question
Lets substitute the values of ${N_1},{N_2},{V_1}$ and ${V_2}$
$ \Rightarrow (0.1 \times 2){V_1} = (0.2 \times 1)50$
We get,
$ \Rightarrow {V_1} = \dfrac{{0.2 \times 50}}{{0.2}}$
$ \Rightarrow {V_1} = 50ml$
Thus, after calculation we know that the correct option is (D)
Note:
Solubility of a substance is the maximum amount of a substance that can be dissolved in a certain amount of solvent. The solubility of a substance depends on the nature of the solute and solvent and the pressure and temperature. It is observed that like dissolves like.
Complete step by step answer:
A solution is a homogenous mixture of two or more components where the composition of the components is the same. There is a solvent which is present in more quantity and there is a solute which is present in lesser quantity as compared to the solvent.
As we know that it is very important to understand the extent and feasibility and application of a solution but that is possible when we know the concentration of each of the components present in the solution. The composition of a solution is determined by the concentration of the components of a solution. There are various methods to calculate the concentration of these components either quantitatively or qualitatively like mass percentage, volume percentage, mole fraction, molarity, normality and others.
For the above given question we use the formula ${N_1}{V_1} = {N_2}{V_2}$.
Where, ${N_1}$ is the normality of the unknown acid or base, ${N_2}$ is the normality of the known base or acid, ${V_1}$ is the volume of unknown acid or base and ${V_2}$ is the volume of the known base or acid. The formula helps to determine the unknown volume or normality for an acid or base with respect to known normality or volume of an acid or base.
On using the formula ${N_1}{V_1} = {N_2}{V_2}$ for the above question
Lets substitute the values of ${N_1},{N_2},{V_1}$ and ${V_2}$
$ \Rightarrow (0.1 \times 2){V_1} = (0.2 \times 1)50$
We get,
$ \Rightarrow {V_1} = \dfrac{{0.2 \times 50}}{{0.2}}$
$ \Rightarrow {V_1} = 50ml$
Thus, after calculation we know that the correct option is (D)
Note:
Solubility of a substance is the maximum amount of a substance that can be dissolved in a certain amount of solvent. The solubility of a substance depends on the nature of the solute and solvent and the pressure and temperature. It is observed that like dissolves like.
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