Question

0.53 gram sodium carbonate dissolved in water and formed 250 ml of solution. calculate the strength, molarity, percentage (w/v) and normality of the solution​

Answer 1

Answer:

Hii

ANSWER

Eq. mass of Na2CO3=2Mol.mass=2106=53

250 mL of the sodium carbonate solution contains =1.325 g

1000 mL of the sodium carbonate solution contains

=2501.325 g×1000=5.300

Normality of Na2CO3 solution =Eq.massStrength(g/L)

=535.30=101N

Applying (Na2CO3)N1V1≡(H2SO4)N2V2

101×25=N2×20

N

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