Question

1.Ten liters of nitric acid of specific gravity of 1.41 containing 67.5 per cent HNO3 by weight were mixed with acid of specific gravity of 1.151 containing 24.99 per cent HNO3 by weight to make an acid of specific gravity of 1.316 containing 50.05 per cent HNO3 by weight. How much of the second acid was used? What is the molarity of the mixture?

Answer 1

We can measure the concentration of a substance in the form of - moles, percentage, weight, volume, mole fraction, and so on.

We know that, Molarity can be calculated by using this formula;

$M = \frac{n}{V}$ ;

where n = Number of moles of the solute

and V = Volume of the solution (per 1 litre)

And, for Mixed Molarity, we use this formula;

$Vc$ × $Mc$ = ($V_1$ × $M_1$) + ($V_2$ × $M_2$)

⇒  $Mc = [(V_1$ × $M_1)$  + $(V_2$ × $M_2)]$ ÷ $Vc$

where,

$Vc =$ Mixed Volume

$Mc$ = Mixed Molarity

$V_1$ = Volume of Solution 1

$M_1$ = Molarity of Solution 1

$V_2$ = Volume of Solution 2

$M_2$ = Molarity of Solution 2

Now, we find the Molarity from a stated gravity and purity (w/w%) via;

M = {Solution's specified gravity × 1000 mL or 1  L× Purity (w/w%) ÷ 100} ÷ Molecular weight

According to the question, we have;

Volume of Nitric Acid = 10 L

Nitric Acid's specified gravity = 1.41

also, containing 67.5% $HNO_3$ by weight;

so, M $(HNO_3)$ = $63 g/mol$

$V_1 = 10 L$

∴ $M_1 =$ (1.41 × 1000 × 67.5 ÷ 100) ÷ 63 $mol/L$ = 15.107

On mixing with the second acid of the specified gravity of 1.151 and having 24.99% $HNO_3$ by weight, we get;

$M_1$ = (1.151 × 1000 × 24.99 ÷ 100) ÷ 63 $mol/L$ = 4.57

Then, the molarity of the mixture;

$(M_m_i_x)$ = 1.316 × 1000 × 50.05 ÷ 100 ÷ 63 = 10.45 $mol/L$

So, mixed molarity of two solutions;

$(M_m_i_x)$ = [($M_1$ × $V_1$) + ($M_2$ × $V_2$)] ÷ ($V_1 + V_2$)

⇒ 10.45 = [(15.107 × 10) + (4.57 × $V_2$)] ÷ (10 + $V_2$)

⇒ 15.107 + 4.57$V_2$ = 10.45 (10 + $V_2$)

⇒ 46.57 = 5.88$V_2$

⇒ $V_2$ = 7.92 L

Ans) Volume of the second acid = 7.92 L; Molarity of the mixture = 10.45 mol/L