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.

$$Noofgmeq=fracWtofSubstance(gm)EqWt$$

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

Molarity

Relationship between normality and molarity:

Percentage (%)

Some Important Formula

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

i) 1.06 g. Na 2 CO 3

We have, g. eqvt.

 = $$fracgivenwteqvtwt$$

= $$frac1.0653=0.02$$

no. of moles

= $$fracgivenwteqvtwt$$

= $$frac1.06106=0.01$$

ii) 150 ml. of 0.1M NaOH solution

We have, g. eqvt. = normality x vol. in litre

= $$0.1\ast frac1501000$$ (in NaOH, N =M)

= 0.015

No. of moles = molarity x vol.in litre

= $$0.1\ast frac1501000$$

= 0.015

2) Calculate the wt. of anhydrous Na2CO3 required to prepare 250 ml. of  $$fracN20(f=1.01)$$ solution.

We have, $$X=fracNEV1000$$

$$=frac1\ast 1.01\ast 53\ast 25020\ast 1000$$

= 0.67 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,$$X=fracNEV1000$$

$$=frac0.25\ast 40\ast 120\ast 10001000$$

= 1200 g.=1.2 kg.

Let the wt. of wet NaOH be y g.

A/Q, (100-12) % of y = x

or, $$frac88100\ast y$$=1.2 kg.

or, y = 1.363 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, V1 = 1L         V2 = (1+x) L

S1 = $$fracM2$$ = 1N

S2 = $$fracN10$$= 0.1 N

(In oxalic acid, N = 2M)

We have, from normality equation:

V1 S1 = V2 S2

or, 1 * 1 = (1 + x)  * $$fracN10$$

or, 1+ x =10

or, x = 9 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

We have, Normality   $$\text{Extra open brace or missing close brace}$$

 $$=frac32\ast 1.16\ast 1036.5$$

= 10.17 N

b) Calculation of vol.

Here, V1 =?     V 2 = 10 L

S1 = 10.17 N     S2 = 1 N

We have, from normality equation:

$$V_1{S}_1=V_2{S}_2$$

or, V 1 x 10.17 N = 10 x 1 N

V1= 0.983 L

6) x g. of oxalic acid reacts completely with 20 ml. of $$fracN10(f=1.06)$$ $$KMn{O}_4$$ solution. Calculate the value of x.

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

or, $$\text{Missing superscript or subscript argument}$$Oxalic acid = ( normality*g. eqvt. of $$KMn{O}_4)$$

$$fracX63=0.1\ast 1.06\ast frac201000$$

X = 63*0.1*1.06*0.02g. 

X = 0.133g. 

7) If 20 ml. of $$fracN20(f=2.06)$$ $$H_{2}S{O}_4$$ and 30 ml.of $$fracN10(f=1.12)$$ HNO 3 are mixed together, calculate the normality of mixture.

We have,  $$V_m{S}_m=V_1{S}_1+V_2{S}_2$$

or, (20 +30) * $$S_m$$ = 20 * 0.05 * 2.06 + 30 * 0.1 * 1.12

or, $$S_m$$ x 50 = 2.06 + 3.36

or, $$S_m$$ = 5.42/50 

∴ $$S_m$$ = 0.1084 N

8) What volume of semi normal and centimolar $$H_{2}S{O}_4$$ solutions should be mixed in order to prepare 1.5 L of decinormal solution of $$H_{2}S{O}_4$$?

Let, $$V_1$$ = y         $$V_2$$ = 1.5-y

$$S_1=fracN2$$

$$S_2=fracM100=fracN50$$

We have,  $$V_m{S}_m=V_1{S}_1+V_2{S}_2$$

or, $$1.5\ast fracN10=y\ast fracN2+(1.5-y)\ast fracN50$$

or, 0.15 = 0.5y + 0.03 – 0.02y

or, 0.48y = 0.12

or, y = $$frac0.120.48$$ = 0.25 L

Thus, the vol. of semi normal $$H_{2}S{O}_4$$= y = 0.25 L

The vol. of centimolar $$H_{2}S{O}_4$$ = 1.5-y = 1.5-0.25=1.25 L

9) 30 c.c. of 0.2N HCl, 10 c.c. of 1N $$H_{2}S{O}_4$$, 20 c.c. of $$fracN10$$HNO3 and 40 c.c. of water are mixed together. What will be the normality of the mixture?

We have, $$V_m{S}_m=V_1{S}_1+V_2{S}_2+V_3{S}_3+V_4{S}_4$$

or, 100 x $$S_m$$ = 30 x 0.2 + 10 x 1 +20 x 0.1 + 40 x 0

or, 100 x $$S_m$$ = 6 + 10 + 2 + 0

or, $$S_m$$ =  $$frac18100$$ = 0.18N

Read: Class 12 Chemistry Notes

Remaining Numericals

Volumetric Analysis Numericals PDF

Read: Volumetric Analysis Note – Class 12 Chemistry