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Question: The volume of water to be added to \[100c{m^3}\]of \[0.5N\] \[{H_2}S{O_4}\] to get decinormal concen...

The volume of water to be added to 100cm3100c{m^3}of 0.5N0.5N H2SO4{H_2}S{O_4} to get decinormal concentration is:
A.100cm3100c{m^3}
B.450cm3450c{m^3}
C.500cm3500c{m^3}
D.400cm3400c{m^3}

Explanation

Solution

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 ,
N1V1=N2V2{N_1}{V_1} = {N_2}{V_2}
Further we are given with the following
N1=0.5N{N_1} = 0.5N
N2=0.1N{N_2} = 0.1N
V1=100cm3{V_1} = 100c{m^3}
V2=?{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
0.5×100=0.1×V20.5 \times 100 = 0.1 \times {V_2}
V2=500cm3{V_2} = 500c{m^3}
Hence, this is the final volume . but we need to find the volume of water to be added to 100cm3100c{m^3}
=500−100= 500 - 100
=400cm3= 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.