You visited us 1 times! Enjoying our articles? Unlock Full Access!

Percentage Composition

Some amount o...

Question

Some amount of $N H_{4} C l$ was boiled with $50 m L$ of $0.75 N N a O H$ solution till the reaction was complete. After the completion of the reaction, $10 m L$ of $0.75 N H_{2} S O_{4}$ were required for the neutralisation of the remaining $N a O H$ . Calculate the amount of $N H_{4} C l$ taken.

Open in App

Solution

$\frac{N_{1} V_{1} + N_{2} V_{2}}{V_{1} + V_{2}} = N \Rightarrow \frac{50 \times 0.75 + 10 \times 0.75 \times 2}{60} = N$
$\frac{37.5 + 15}{60} = N = \frac{52.5}{60} = 0.875 N$
$\frac{A m o u n t o f N H_{4} C l}{m o l a r m a s s o f N H_{4} C l} = N \times v o l . \Rightarrow 0.875 \times 0.06 = 0.00525$
$A m o u n t o f N H_{4} C l = 53.4 \times 0.00525 = 0.27025 g .$

Suggest Corrections

Similar questions

100 m L

0.8 N N a O H

N a O H

12.5 m L

0.75 N H_{2} S O_{4}

H_{2} C_{2} O_{4}

N a H C_{2} O_{4}

2.02 g

10 m L

3.0 m L

0.1 N N a O H

10 m L

H_{2} S O_{4}

4 m L

0.1 N K M n O_{4}

H_{2} C_{2} O_{4}

N a H C_{2} O_{4}

2.26 g

100 m L

N N a O H

N a O H

30 m L 2 N H_{2} S O_{4}

H_{2} S O_{4}

H_{2} C_{2} O_{4}

3.185 g

1 l i t r e . 10 m L

3 m L

0.1 N

N a O H

100 m L

4 m L

0.02 M K M n O_{4}

%

H_{2} S O_{4}

Join BYJU'S Learning Program