Question

The normality of a solution that results from mixing $4g$ of $NaOH$, $500ml$ of $1M$ $HCl,$ and $10.0ml$ of ${H_2}{O_4}$ (specific gravity $1.1,49\% {H_2}S{O_4}$ by weight ) is : (The total volume of the solution was made to $1L$ with water )
A. $0.51$
B. $0.71$
C. $1.02$
D. $0.45$

Answer 1

Hint: We will approach this problem with the help of formula of normality which is given below;
Normality of ${H_2}S{O_4}$ $= \dfrac{{{W_2} \times 1000}}{{{E_{w2}} \times {V_{sol}}(inml)}} = \dfrac{{\% weight \times 10 \times d}}{{{E_{w2}}}}$, where ${W_2}$, ${E_{w2}}$ are weight in gram and equivalent weight of second substance that is Sulphuric acid.

Complete answer:
We know the weight of $NaOH$ in grams is $4g$ so we can calculate the number of moles of $NaOH$.
Number of moles of Sodium hydroxide =weight of sodium hydroxide in gram/molecular weight of sodium hydroxide;( Molecular weight of $NaOH$ = atomic weight of Sodium ($Na$)+ atomic weight of oxygen ($O$) +atomic weight of Hydrogen ($H$) = $23 + 16 + 1 = 40$ ).
So number of moles of $NaOH$ $= \dfrac{4}{{40}} = 0.1mole$ ; $0.1mole$ is equal to $0.1mmol$or we can say that it is equal to $100mEq$.Now we know the milli equivalent of $NaOH$ .Lets calculate milli equivalent of ${H_2}S{O_4}$ with the help of formula given in the hint.
Normality of ${H_2}S{O_4}$ $= \dfrac{{{W_2} \times 1000}}{{{E_{w2}} \times {V_{sol}}(inml)}} = \dfrac{{\% weight \times 10 \times d}}{{{E_{w2}}}}$
Equivalent weight of sulphuric acid can be calculated as the ratio of molecular weight to basicity ,so equivalent weight of ${H_2}S{O_4}$ $= \dfrac{{98}}{2} = 49$
Percentage by weight of ${H_2}S{O_4}$ is 49 and the density (specific gravity ) is $1.1$ so putting the value of these in the above equation we get;
Normality of ${H_2}S{O_4}$ $= \dfrac{{49 \times 10 \times 1.1}}{{49}} = 1.1N$
So milli equivalent of ${H_2}S{O_4}$ $= 11N \times 10.0mL = 110mEq$
Total acid =$110 + 500 = 610mEq$ and we have calculated milli equivalent of $NaOH$ $= 100mEq$
So the acid left $= 610 - 100 = 510mEq$
Total volume is one litre so the normality of solution $= \dfrac{{mEq}}{{mL}} = \dfrac{{510}}{{1000}} = 0.51N$
Here option A is correct that is $0.51N$

Note: In moles and normality problems we can solve questions with the help of some important formula. As here in the problem many values are given so we just need to put them in the formula and need to find out milli equivalent of $NaOH$ and then we also find out milli equivalent of ${H_2}S{O_4}$ . At last we have found out the milli equivalent of the left acid and the volume of the solution is one litre then with the formula normality of solution $= \dfrac{{mEq}}{{mL}}$ we have found out the normality of the solution.