
The volume of water to be added to \[100c{m^3}\]of \[0.5N\] \[{H_2}S{O_4}\] to get decinormal concentration is:
A.\[100c{m^3}\]
B.\[450c{m^3}\]
C.\[500c{m^3}\]
D.\[400c{m^3}\]
Answer
547.8k+ views
Hint:We are already given with the above quantities and according to the question we need to check which equation is related to the topic of finding volume which should be added to get the decinormal concentration.
Complete step by step answer:
Normality: the number of grams or mole equivalent of solute into one litre of solution.
Volume : it is the quantity of three dimensional space which is occupied by solid , liquid or gas. The unit of volume is cubic metre.
As here in the question we have given normality , volume of the solution so we can use the relation of normality and volume .
When we talk about the concentration we come to know that it is contribution by any substance that is making a part of solvent. It is the Amount Of substance in a defined space .
Decinormal concentration defines concentration having one tenth of a normal solution.
As we know that ,
\[{N_1}{V_1} = {N_2}{V_2}\]
Further we are given with the following
\[{N_1} = 0.5N\]
\[{N_2} = 0.1N\]
\[{V_1} = 100c{m^3}\]
\[{V_2} = ?\]
From this we concluded that we need to find the final volume .
Now putting the given quantities in the above equation we found that
\[0.5 \times 100 = 0.1 \times {V_2}\]
\[{V_2} = 500c{m^3}\]
Hence, this is the final volume . but we need to find the volume of water to be added to \[100c{m^3}\]
\[ = 500 - 100\]
\[ = 400c{m^3}\]
Hence the correct option is (D).
Note:
Deci normal solution : When one-tenth gram equivalent mass of a substance is present in one litre of its solution then it is called decinormal solution . it is denoted by N/10 solution.
Complete step by step answer:
Normality: the number of grams or mole equivalent of solute into one litre of solution.
Volume : it is the quantity of three dimensional space which is occupied by solid , liquid or gas. The unit of volume is cubic metre.
As here in the question we have given normality , volume of the solution so we can use the relation of normality and volume .
When we talk about the concentration we come to know that it is contribution by any substance that is making a part of solvent. It is the Amount Of substance in a defined space .
Decinormal concentration defines concentration having one tenth of a normal solution.
As we know that ,
\[{N_1}{V_1} = {N_2}{V_2}\]
Further we are given with the following
\[{N_1} = 0.5N\]
\[{N_2} = 0.1N\]
\[{V_1} = 100c{m^3}\]
\[{V_2} = ?\]
From this we concluded that we need to find the final volume .
Now putting the given quantities in the above equation we found that
\[0.5 \times 100 = 0.1 \times {V_2}\]
\[{V_2} = 500c{m^3}\]
Hence, this is the final volume . but we need to find the volume of water to be added to \[100c{m^3}\]
\[ = 500 - 100\]
\[ = 400c{m^3}\]
Hence the correct option is (D).
Note:
Deci normal solution : When one-tenth gram equivalent mass of a substance is present in one litre of its solution then it is called decinormal solution . it is denoted by N/10 solution.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

