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Standard XII

Chemistry

Stoichiometric Calculations

30 mL of N/...

Question

$30 m L$ of $N / 10 H C l$ is required to neutralise $50 m L$ of a sodium carbonate solution. How many $m L$ of water must be added to $30 m L$ of this solution so that the solution obtained may have a concentration equal to $N / 50$ ?

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Solution

Given,
$V_{H C l} = 30 m l$ , $N_{H C l} = \frac{N}{10} = 0.1 N$
$N_{N a_{2} C O_{3}} = ?$ $V_{N a_{2} C O_{3}} = 50 m L$

$(V \times N)_{N a_{2} C O_{3}} = (V \times N)_{H C l}$
$N_{N a_{2} C O_{3}} = \frac{30 \times 0.1}{50}$
$= 0.06 N$
Again, $30 m L$ of $0.06 N$ $N a_{2} C O_{3}$ is made to $\frac{N}{50} = 0.02 N$
$N_{1} V_{1} = N_{2} V_{2}$
$V_{2} = \frac{N_{1} V_{1}}{N_{2}} = \frac{0.06 \times 30}{0.02} = 90$ mL$
$90 m L$ of water should be zadded to get $0.02 N$ solution.

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30 mL of N/10 HCl is required to neutralise 50 mL of a sodium carbonate solution. How many mL of water must be added to 30 mL of this solution so that the solution obtained may have a concentration equal to N/50?

Given,VHCl=30ml, NHCl=N10=0.1NNNa2CO3=? VNa2CO3=50mL(V×N)Na2CO3=(V×N)HCl NNa2CO3=30×0.150 =0.06NAgain, 30mL of 0.06N Na2CO3 is made to N50=0.02NN1V1=N2V2V2=N1V1N2=0.06×300.02=90 mL$90mL of water should be zadded to get 0.02N solution.

$30 m L$ of $N / 10 H C l$ is required to neutralise $50 m L$ of a sodium carbonate solution. How many $m L$ of water must be added to $30 m L$ of this solution so that the solution obtained may have a concentration equal to $N / 50$ ?