Question

50 ⓟⓞⓘⓝⓣⓢ!!

p̾l̾z̾ n̾h̾i̾ a̾a̾t̾a̾ t̾o̾ d̾u̾r̾ r̾h̾o̾ ✋​

Answer 1

Hey !!

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→ First of all find equivalent weight of Na2CO3

equivalent weight = molecular weight ÷ valency factor

equivalent wt = 106 /2 =53g

now,

find normality

given wt/ equivalent wt. ÷ volume in L

10.6 /53 ÷ (100/1000)

=2N

hope it helps!!

#jai hind

Answer 2

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

10.6 kg of Na²CO³ was exactly neutralized by 100ml of H²80⁴

$\rule{200}3$

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

→ First of all find equivalent weight of Na2CO3

equivalent weight = molecular weight ÷ valency factor

equivalent wt = 106 /2 =53g

now,

find normality

given wt/ equivalent wt. ÷ volume in L

10.6 /53 ÷ (100/1000)

=2N

$\rule{200}3$

# own composition✔