Question

Find the amount of Na2CO3 required to prepare 250 ml of 0.1 N sodium carbonate solution? *
.

Answer 1

Answer:

1.325 g

Explanation:

Normality = 0.1  N

V = 250 ml = 0.25 L

Normality = no. of equivalents[$n_{eq}$] / volume[in L]

=> 0.1 N = $n_{eq}$ x 0.25

=> $n_{eq}$ = 0.025

Number of equivalents[$n_{eq}$] = given mass[w] / equivalent mass

=> 0.025 = w / 53

[Eqlnt mass of Na2CO3 = molar mass/n-factor = 106/2 = 53]

=> w = 0.025 x 53 = 1.325 g

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Answer 2

Answer:

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