You visited us 8 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Mole Concept

1.575 g of ox...

Question

$1.575 g$ of oxalic acid $(C O O H)_{2} \cdot x H_{2} O$ are dissolved in water and the volume made up to $250 m L$ . On titration $16.68 m L$ of this solution requires $26 m L$ of $\frac{N}{15} N a O H$ solution for complete neutralisation. Calculate $x$ .

Open in App

Solution

$25 m L$ of $\frac{N}{15} N a O H$ solution $\equiv 25 m L$ of $\frac{N}{15}$ oxalic acid solution
Mass of oxalic acid present in $25 m L$ of $\frac{N}{15}$ oxalic acid solution
$= \frac{N \times E \times V}{1000} = \frac{1 \times (90 + 18 x) \times 25}{15 \times 2 \times 1000}$
$= \frac{(90 + 18 x)}{1200} g$
Actually $\frac{(90 + 18 x)}{1200} g$ oxalic acid is present in $16.68 m L$ solution.
$250 m L$ of the solution contains oxalic acid
$= \frac{(90 + 18 x) \times 250}{1200 \times 16.68} = 1.575 (g i v e n)$
or $90 + 18 x = \frac{1.575 \times 1200 \times 16.68}{250} = 126$
or $18 x = 126 - 90 = 36$
$x = 2$ .

Suggest Corrections

Similar questions

4 g

of a mixture of

N a C l

and

N a_{2} C O_{3}

were dissolved in water and volume made up to

250 m L

15 m L

of this solution required

50 m L

N / 10 H C l

for complete neutralisation. Calculate the percentage composition of the original mixture.

4 g

of a mixture of

N a_{2} S O_{4}

and anhydrous

N a_{2} C O_{3}

were dissolved in pure water and volume made up to

250 m L

20 m L

of this solution required

25 m L

N / 5 H_{2} S O_{4}

for complete neutralisation. Calculate the percentage composition of the mixture.

Q. An aqueous solution of

6.3 g

of oxalic acid dihydrate is made up to

250 m L

. The volume of

0.1 N N a O H

required to completely neutralise

10 m L

of this solution is :

0.7 g of Na₂Co₃.XH₂O were dissolved in water and the volume was made to 100mL , 20 mL of this solution required 19.8 mL of N/10 HCl for complete neutralisation, the value of 'x' is

A) 7

B) 3

C) 2

D) 5

1.325 g

of anhydrous sodium carbonate are dissolved in water and the solution made up to

250 m L

. On titration

25 m L

of this solution neutralise

20 m L

of a solution of sulphuric acid. How much water should be added to

450 m L

of this acid solution to make it exactly

N / 12

Join BYJU'S Learning Program

1.575 g of oxalic acid (COOH)2⋅xH2O are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this solution requires 26 mL of N15NaOH solution for complete neutralisation. Calculate x.

$1.575 g$ of oxalic acid $(C O O H)_{2} \cdot x H_{2} O$ are dissolved in water and the volume made up to $250 m L$ . On titration $16.68 m L$ of this solution requires $26 m L$ of $\frac{N}{15} N a O H$ solution for complete neutralisation. Calculate $x$ .