The concentration of solutions can be expressed in number of ways such that Normality, Molarity, Molality, Mole fractions, Strength, `%` by weight, `%` by volume and `%` by strength. The molarity of ionic compound is usually expressed as formality beacuse we use formula weight of ionic compound. Addition of water to a solution changes all these terms, however increase in temperature does not change molality, mole fraction and `%` by weight terms.
Two litre of `NH_(3)` at `30^(@)C` and `0.20 atm` is neutralised by `134 mL` of acid `(H_(2)SO_(4))`. The molarity of `H_(2)SO_(4)` is:

Options

A. `0.12`
B. `0.24`
C. `0.06`
D. `0.03`

Answer

B

Explanation

For `NH_(3), PV=(w)/(m)RT`
`therefore (w)/(m)=(PV)/(RT)=(0.2xx2)/(0.0821xx303)=0.01608`
`therefore` Mole of `NH_(3)= "Equivalent of" NH_(3)`
`=0.01608`
`therefore Meq.of NH_(3)=16.08`
Now Meq. of `H_(2)SO_(4)="Meq.of" NH_(3)`
`Nxx134=16.08`
`therefore `N=0.12`