Question

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Answer 1

hello Grandpa max!!

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First of all find

→ Molecular weight of Na2CO3.

= 23 × 2 + 12 + 48

=106g

→ equivalent weight = Molecular weight/Valency factor

= 106/2 =53

→ Normality = given wt./ (g.equivalent wt × Volume in litre)

= 10.6 × 1000 /(53×100)

=106/53

=2N

hope you get your answer!!

#JaiHind

Answer 2

$\huge\frak{\underline{\underline{Question:}}}$

◾️10.6g of Na2C03 was exactly neutralised by 100 ml of H2 S04 solution. Its normally is?

$\huge\frak{\underline{\underline{Answer:}}}$

$\Huge{\boxed{\boxed{\bold{20N}}}}$

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Explanation:

Molecular weight of Na2CO3.

$\implies$ $\bold{23 × 2 + 12 + 48}$

$\implies$ $\bold{106g}$

$\implies$$\bold{equivalent\:weight\:=\frac{Molecular\:weight}{Valency\:factor}}$

$\implies$ $\bold{\frac{106}{2}} =53$

$\implies$$\bold{Normality =\frac{given wt.}{g.equivalent wt × Volume in litre}}$

$\implies$ $\bold{\frac{10.6 × 1000}{53×100}}$

$\implies$$\bold{\frac{106}{53}}$

$\implies$ $\bold{2N}$

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