You visited us 3 times! Enjoying our articles? Unlock Full Access!

Ways of Representing the Concentration of a Solution

A sample of s...

Question

A sample of sodium carbonate contain sodium sulphate also. $1.5 g$ of the sample is dissolved in water and volume raised to $250 m L$ . $25 m L$ of this solution requires $20 m L$ of $\frac{N}{10} H_{2} S O_{4}$ solution for neutralisation. Calculate percentage of sodium carbonate in the sample.

Open in App

Solution

Only $N a_{2} C O_{3}$ will react with $H_{2} S O_{4}$ .
Applying $N_{1} V_{1} (N a_{2} C O_{3}) = N_{2} V_{2} (H_{2} S O_{4})$
$N_{1} \times 25 = 20 \times \frac{1}{10}$
$N_{1} = \frac{20}{25 \times 10} = 0.08$
Eq. mass of $N a_{2} C O_{3} = \frac{M o l . m a s s}{2} = \frac{106}{2} = 53$
Mass of $N a_{2} C O_{3}$ present in $250 m L 0.08 N$ solution
$= \frac{N \times E \times V}{1000} = \frac{0.08 \times 53 \times 250}{1000} = 1.06 g$ $
Percentage of $N a_{2} C O_{3}$ in the mixture $= \frac{1.06}{1.50} \times 100 = 70.67$ .

Suggest Corrections

Similar questions

1.25 g

250 m L

25 m L

20 m L

\frac{N}{10}

1.5 g

500 m L

25 m L

24 m L

26 m L

25 m L

N / 20

12 g

7.5 g

250 m L

25 m L

22.4 m L

25 m L

24.8 g

(N a_{2} S_{2} O_{3} \cdot 5 H_{2} O)

A s = 75)

A

2.0

C O_{2}

C O_{2}

750

298

123.9

A

1.5

150

\frac{M}{10} H C l

1.25 g

{N a}_{2} {C O}_{3}

{N a}_{2} {S O}_{4}

250 m l

25 m l

20 m l

0.1 N

H_{2} {S O}_{4}

{N a}_{2} {C O}_{3}

Join BYJU'S Learning Program