## 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}`**

Compounds | Molecular wt. | Equivalent wt. | n |
---|---|---|---|

Hydrochloric acid | 36.5 | 36.5 | 1 |

Sulphuric acid | 98 | 49 | 2 |

Oxalic acid | 126 | 63 | 2 |

Sodium hydroxide | 40 | 40 | 1 |

Sodium carbonate | 106 | 53 | 2 |

Calcium carbonate | 100 | 50 | 2 |

Aluminium | 27 | 9 | 3 |

Magnesium | 24 | 12 | 2 |

**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

### Gm/liter:

### 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.

= `frac{given wt}{eqvtwt} `

= `\frac{1.06}{53} = 0.02`

no. of moles

= `frac{given wt}{eqvtwt} `

= `\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 `\frac{N}{20}(f = 1.01)` solution.

We have, `X = \frac{NEV}{1000}`

`= \frac{1*1.01*53*250}{20*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 = \frac{NEV}{1000}`

`= \frac{0.25*40*120*1000}{1000}`

= 1200 g.=1.2 kg.

Let the wt. of wet NaOH be y g.

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

or, `\frac{88}{100}*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 = `\frac{M}{2}` = 1N

S2 = `\frac{N}{10} `= 0.1 N

(In oxalic acid, N = 2M)

We have, from normality equation:

V1 S1 = V2 S2

or, 1 * 1 = (1 + x) * `\frac{N}{10} `

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 `= \frac{\% (\frac{w}{w})*sp. gravity*10}{eqvt wt\ldotp }`

`= \frac{32*1.16*10}{36.5}`

= 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 `\frac{N}{10}(f = 1.06)` ` KMnO_4` solution. Calculate the value of x.

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

or, `(\frac{given wt.}{eq. wt})_`Oxalic acid = ( normality*g. eqvt. of `KMnO_4)`

`\frac{X}{63} = 0.1*1.06*\frac{20}{1000}`

X = 63*0.1*1.06*0.02g.

X = 0.133g.

### 7) If 20 ml. of `\frac{N}{20}(f = 2.06)` `H_2SO_4` and 30 ml.of `\frac{N}{10}(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_2SO_4` solutions should be mixed in order to prepare 1.5 L of decinormal solution of `H_2SO_4`?

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

`S_1 = \frac{N}{2}`

`S_2 = \frac{M}{100} = \frac{N}{50}`

We have, `V_m S_m = V_1 S_1 + V_2 S_2`

or, `1.5 * \frac{N}{10} = y * \frac{N}{2} + (1.5-y) * \frac{N}{50}`

or, 0.15 = 0.5y + 0.03 – 0.02y

or, 0.48y = 0.12

or, y = `\frac{0.12}{0.48}` = 0.25 L

Thus, the vol. of semi normal `H_2SO_4 `= y = 0.25 L

The vol. of centimolar `H_2SO_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_2SO_4 `, 20 c.c. of `\frac{N}{10}`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` = `\frac{18}{100}` = 0.18N

**Read**: Class 12 Chemistry Notes

### Remaining Numericals

## Volumetric Analysis Numericals PDF

Read: Volumetric Analysis Note - Class 12 Chemistry