Ten liters of nitric acid of specific gravity of 1.41 containing 67.5% HNO3 by weight were mixed with acid of specific gravity of 1.151 containing 24.99% HNO3 by weight to make an acid of specific gravity of 1.316 containing50.05% HNO3 by weight. How much of the second acid was used? What is the molarity of the mixture?​

Answers

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Explanation:

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 + 4.57V_2 = 10.45 (10 + V_2)

⇒ 46.57 = 5.88V_2

V_2 = 7.92 L

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



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