Volumetric Analysis Numericals Problems – Class 12 Chemistry

Volumetric Analysis Numericals Problem Solution

Grade 12 Chemistry Chapter 1 Volumetric Analysis Numericals Questions Answer Solution

Normality

Normality is defined as the no. of gram equivalent of solute present in one liter of the solution. It is represented as N.

No. of gm eq. = V of Solution (in L) * Normality

$$No. of gm eq. = \frac{V*N}{1000}$$

$$\text{Normality} = \frac{\text{Wt of solute in gm}}{\text{Eq wt * V of Solution (in L)}}$$

$$\frac{\text{Wt of solute in gm}}{\text{Eq wt} * \frac{\text{V of Solution (in mL)}}{1000}}$$

$$\text{Wt. of Solute in gm} = \frac{\text{Normality * Eq wt * V of Solute (in ml)}}{1000}$$

$$\therefore w = \frac{VEN}{1000}$$

The solution obtained by dissolving one gram equivalent weight of solute in one liter of solution is called normal solution.

$$No of gm eq = \frac{\text{Wt of Substance (gm)}}{\text{Eq Wt}}$$

Similarly, the solution obtained by dissolving 1/10th gram equivalent weight of solute in one liter of solution is called decinormal solution.

How many gram of sodium carbonate is required to prepare 250ml of decinormal solution of sodium carbonate?

Calculate the concentration of solution obtained by dissolving 0.63gm of oxalic acid into 200ml of solution.

Molarity

Molarity of a solution may be defined as the no. of moles of solute present in one liter of the solution. It is represented as M.

$$No of Moles = \frac{\text{Wt of Substance (gm)}}{\text{Mol. Wt}}$$

Similarly, the solution obtained by dissolving 1/10th mole of solute in one liter of solution is called decimolar solution.

Gm/liter:

The amount of solute in gram that is present in one liter or 1000ml of solution is called gram per liter.

Relationship between normality and molarity:

Percentage (%)

The amount of solute in gram that is present in 100ml of solution is called percentage (by volume).

Some Important Formula

1) Express the following in g. eqvt. and moles

i) 1.06 g. Na₂CO₃

We have, g. eqvt. = $$\frac{\text{given wt}}{\text{eqvt wt}} = \frac{1.06}{53} = 0.02$$

no. of moles = $$\frac{\text{given wt}}{\text{mol wt}} = \frac{1.06}{106} = 0.01$$

ii) 150 ml. of 0.1M NaOH solution

We have, g. eqvt. = normality x vol. in litre = $$0.1 * \frac{150}{1000}$$ (In NaOH, N = M)

= 0.015

No. of moles = molarity x vol. in litre = $$0.1 * \frac{150}{1000} = 0.015$$

2) Calculate the wt. of anhydrous Na₂CO₃ required to prepare 250 ml. of N/20 (f = 1.01) solution.

We have, $$X = \frac{NEV}{1000} = \frac{1 * 1.01 * 53 * 250}{20 * 1000} = 0.67 \text{g}$$

3) How many kg of wet NaOH containing 12% water are required to make 120 L of a 0.25 N solution.

We have, $$X = \frac{NEV}{1000} = \frac{0.25 * 40 * 120 * 1000}{1000} = 1200 \text{g} = 1.2 \text{kg}$$

Let the wt. of wet NaOH be y g. A/Q, (100-12) % of y = x

or, $$\frac{88}{100} * y = 1.2 \text{kg} \implies y = 1.363 \text{kg}$$

4) You are given a L of semi molar solution of oxalic acid. What volume of water should be added into it to make it exactly decinormal?

Here, $V_1 = 1L$, $V_2 = (1+x) L$, $S_1 = M/2 = 1N$, $S_2 = N/10 = 0.1 N$ (In oxalic acid, N = 2M)

From normality equation: $V_1 S_1 = V_2 S_2$

or, $1 * 1 = (1 + x) * 0.1 \implies 1+ x = 10 \implies x = 9 \text{L}$

5) Concentrated hydrochloric acid has a specific gravity of 1.16 & contains 32% HCl by wt. Calculate the vol. of this acid which would be required to make 10 L of a normal solution of the acid.

a) Calculation of concentration of conc. HCl

Normality = $$\frac{\% (w/w) * \text{sp. gravity} * 10}{\text{eqvt wt}} = \frac{32 * 1.16 * 10}{36.5} = 10.17 N$$

b) Calculation of vol.

Here, $V_1 = ?, V_2 = 10 L, S_1 = 10.17 N, S_2 = 1 N$

Normality equation: $V_1 S_1 = V_2 S_2 \implies V_1 * 10.17 = 10 * 1 \implies V_1 = 0.983 L$

6) x g. of oxalic acid reacts completely with 20 ml. of N/10 (f = 1.06) KMnO₄ solution. Calculate the value of x.

g. eqvt. of oxalic acid = g. eqvt. of KMnO₄

$$\frac{x}{63} = 0.1 * 1.06 * \frac{20}{1000} \implies x = 63 * 0.1 * 1.06 * 0.02 = 0.133\text{g}$$

7) If 20 ml. of N/20 (f = 2.06) H₂SO₄ and 30 ml. of N/10 (f = 1.12) HNO₃ are mixed together, calculate the normality of mixture.

$$V_m S_m = V_1 S_1 + V_2 S_2$$

or, $(20 + 30) * S_m = 20 * 0.05 * 2.06 + 30 * 0.1 * 1.12 \implies S_m * 50 = 2.06 + 3.36 \implies S_m = 0.1084 N$$

8) What volume of semi normal and centimolar H₂SO₄ solutions should be mixed in order to prepare 1.5 L of decinormal solution of H₂SO₄?

Let $V_1 = y$, $V_2 = 1.5 – y$. $S_1 = N/2$, $S_2 = M/100 = N/50$

$$1.5 * \frac{1}{10} = y * \frac{1}{2} + (1.5 – y) * \frac{1}{50} \implies 0.15 = 0.5y + 0.03 – 0.02y \implies 0.48y = 0.12 \implies y = 0.25 L$$

Thus, vol. of semi normal = 0.25 L, vol. of centimolar = 1.25 L

9) 30 c.c. of 0.2N HCl, 10 c.c. of 1N H₂SO₄, 20 c.c. of N/10 HNO₃ and 40 c.c. of water are mixed together. What will be the normality of the mixture?

$$V_m S_m = V_1 S_1 + V_2 S_2 + V_3 S_3 + V_4 S_4$$

or, $100 * S_m = 30 * 0.2 + 10 * 1 + 20 * 0.1 + 40 * 0 \implies 100 * S_m = 18 \implies S_m = 0.18 N$$

Read: Class 12 Chemistry Notes

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Volumetric Analysis Numericals PDF

Read: Volumetric Analysis Note – Class 12 Chemistry