1.

How manymillilitres of concentrated sulphuric acidof sp. 1.84 containing 98 %H_(2)SO_(4) by weightare required to prepare 200 mLof 0.50 N solution ?

Answer»

Solution :98 % of `H_(2)SO_(4)`by weightmeans100 g `H_(2)SO_(4)`solutioncontains98 gof `H_(2)SO_(4)`
Volume of 100g of `H_(2)SO_(4) = 100/(1.84 ) mL `
i.e `100/(1.84)` mL of `H_(2)SO_(4)` solution contains98.00g of `H_(2)SO_(4)`
Equivalents of `H_(2)SO_(4) = 98/49 = 2 ""...(EQN .4i)`
(eq. WT . of `H_(2)SO_(4)=49`)
m.e of `H_(2)SO_(4) = 2xx1000 = 2000. "" ...(Eqn .3)`
Normality of `H_(2)SO_(4)` solution = `(m.e)/("volume in mL") "" ....(Eqn.1)`
` = (2000)/(100//1.84) = 36.8 N `
Let the volume of `H_(2)SO_(4)`of normality36.8 N = 36.8 V ....(Eqn.1)
and m.e of 200 mL of `H_(2)SO_(4)` of normality `0.5 N = 0.5 xx 200`
= 100
Since both the solutions of `H_(2)SO_(4)`shouldhave the same number of m.e we have ,
` 36.8 v = 100`
` :. "" v = 100/(36.8) = 2.72 mL `


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