To 50 mL of 0.1 N Na_2CO_3 solution,150 m | vedclass

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(C) Initial volume $(V_1) = 50 \ mL$Initial normality $(N_1) = 0.1 \ N$Final volume $(V_2) = 50 \ mL + 150 \ mL = 200 \ mL$Using the dilution formula:

Vedclass · Accepted Answer

$M/80$

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(C) Initial volume $(V_1) = 50 \ mL$Initial normality $(N_1) = 0.1 \ N$Final volume $(V_2) = 50 \ mL + 150 \ mL = 200 \ mL$Using the dilution formula: $N_1 	imes V_1 = N_2 	imes V_2$$0.1 	imes 50 = N_2 	imes 200$$N_2 = \frac{0.1 	imes 50}{200} = 0.025 \ N = \frac{1}{40} \ N$For $Na_2CO_3$,the valence factor $(Z)$ is $2$ (total positive charge of $Na^+$ ions).Relation between Normality and Molarity: $N = M 	imes Z$$M = \frac{N}{Z} = \frac{1/40}{2} = \frac{1}{80} \ M$Therefore,the molarity of the resultant solution is $M/80$.

To $50 \ mL$ of $0.1 \ N \ Na_2CO_3$ solution,$150 \ mL$ of water is added. What is the molarity of the resultant solution?

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