Question

The normality of 0.05 m H2SO4 solution will be?

Answer 1

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } CorrecT\: Question:-}}}}}$

The normality of 0.05M $\sf H_2SO_4$solution will be?

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } SolutioN:-}}}}}$

Given that Molarity of $H_2SO_4$ solution is 0.05 M .

So we can find Normality as ,

$\large\boxed{\purple{\tt \red{\dag} Normality\:(N)\:=\:n-factor\:\times\: Molarity}}$

On mixing water , it disassociates to give 2 $\sf H^{+}$ ions .

$\boxed{\red{\blue{\bullet}\:\bf \:H_2SO_4\:\longrightarrow 2H^{+}\:+\:SO_4^{2-}}}$

Now n - factor of Sulphuric acid or it's basicity is 2 , and molarity is 0.05 M .

$\tt:\implies Normality=n-factor\times Molarity$

$\tt:\implies N =2\times0.05$

$\tt:\implies N=\cancel{2}\times\dfrac{\cancel{5}}{\cancel{100}}$

$\underline{\boxed{\red{\tt \longmapsto \:\:Normality\:\:=\:\:0.1}}}$

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$\underline{\pink{\leadsto \mathscr MORE\: INFORMATION:}}$

$\underline{\green{\bf\bullet Molarity:-}}$

Number of moles of solute per litre of solution is known as Molarity. It is denoted by M .

$\underline{\green{\bf\bullet Molality:-}}$

Number of moles of solute per kg of solvent is known as molality. it is denoted by m .

$\underline{\green{\bf\bullet Normality:-}}$

Gram equivalent weight of solute by volume of solution in litre is known as normality. where gram equivalent weight is equal to equivalent weight by valency.