Question: 0.45 N and 0.6 N solutions are added in 2 : 1 by volume. the amount of solute present in one litre o...

Question

0.45 N and 0.6 N solutions are added in 2 : 1 by volume. the amount of solute present in one litre of this solution is
A. 0.5gm
B. 25 gm.
C. 20 gm.
D. 5 gm

Answer 1

Solution

To get the answer of these types of questions we have to firstly calculate the overall normality of the solution by mixing both solution and final normality of the solution. After calculating the normality of the whole solution we can find the number of gram equivalents of the compound and finally at last we can find the mass of the solute..

Complete solution:
Now we have two types of solution:
First solution is of 0.45 normality
And the second solution is 0.6 normality.
After combining the both solution, and finding the normality of solution
We will use following formula:
${N_{mixture}} = \dfrac{{{N_1}{V_1} + {N_2}{V_2}}}{{{N_1} + {N_2}}}$
Where, ${N_1}$ is the normality of first solution
${N_2}$ is the normality of the second solution
${V_1}$ is the volume of first solution
${V_2}$ is the volume of the second solution
Now after putting the all values in above given formula,
We get
$\dfrac{{(0.45 \times 2) + 0.6}}{{2 + 1}} = \dfrac{{0.9 + 0.6}}{3} = \dfrac{{1.5}}{3} = 0.5$
( $\therefore$ we first solution is used with 2 unit volume and second solution is used with 1 unit volume,
Hence, we have multiplied by 2 in the first solution.)
Now we have calculated the normality of mixture is 0.5 N i.e. 0.5 gram equivalent of solute is used in 1 litre of the solution
Hence, we can say that 0.5 gram of solute is used with per equivalent mass
Therefore, finally if acidity or basicity of solute is 1 then the total amount of solute will be 0.5 gram in mixture.

So, correct answer will be option number A.

Note: Normality is defined as the number of gram equivalents of solute present per litre of solution. it is denoted by ‘N’. Mathematically normality is equal to number of gram equivalents of solute upon number of litres of solutionGram equivalent it is defined as the ratio of given mass of the solute divided by equivalent mass of that soluteEquivalent mass is defined as the ratio of molar mass of that substance upon acidity or basicity of that substance.