Question

A solution of H2so4 is 80percent by wt, having specific gravity 1.73 it's normality is??

Answer 1

$\huge\mathbb{\green{QUESTION-}}$

A solution of H2so4 is 80percent by wt, having specific gravity 1.73 it's normality is

$\huge\mathbb{\green{SOLUTION-}}$

$\large\underline{\underline{\sf Given:}}$

• ${\sf H_2SO_4}$ is 80% by weight
• Specific Gravity = 1.73

$\large\underline{\underline{\sf To\:Find:}}$

• Normality (N)

$\large\underline{\underline{\sf Formula\:Used:}}$

$\large{\boxed{\sf Normality (N)=\dfrac{No.\;of\:gram\: Equivalent}{Volume\:of\: Solution (in\:litre)}}}$

${\sf H_2SO_4}$ = 80% by weight

100g = 80g ${\sf H_2SO_4}$

$\large{\sf \red{No.\:of\:gram\: Equivalent=\dfrac{Mass}{Equivalent\: weight}}}$

$\large{\sf \red{Equivalent\: Weight (E)=\dfrac{Molecular\: Weight}{n-factor}} }$

Molecular Weight of ${\sf H_2SO_4}$ is ⎯

2 × 1 + 32 + 4 × 16

98 g

n - factor = 2

${\sf No.\:of\: gram\: Equivalent=\dfrac{mass}{Equivalent\; Weight} }$

$\implies{\sf \dfrac{80}{\dfrac{98}{2}}}$

$\implies{\sf \dfrac{80}{49}}$

Specific gravity = 1.71

density of Solution = 1.71 g/mL

Volume of Solution = ${\sf \dfrac{100}{1.71}}$mL

${\sf \blue{Normality=\dfrac{No.\:of\: Equivalent\:of\: Solute}{Volume\:of\: Solution\:(in\:litre)} }}$

$\implies{\sf \dfrac{80}{49}×\dfrac{1.71}{10^{-1}} }$

$\implies{\bf 27.9\:N}$

$\huge\mathbb{\green{ANSWER-}}$

Normality is ${\bf \red{27.9\:N}}$