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in Redox reactions and electrochemistry by (15 points)
4. \( 1.20 g \) sample of \( Na _{2} CO _{3} \) and \( K _{2} CO _{3} \) was dissolved in water to form \( 100 mL \) of a Solution. \( 20 mL \) of this solution required \( 40 mL \) of \( 0.1 N HCl \) for complete neutralization. Calculate the weight of \( Na _{2} CO _{3} \) in the mixture. If another \( 20 mL \) of this solution is treated with excess of \( BaCl _{2} \) what will be the weight of the precipitate?

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Given-

Mass of sample = 1.20g

Normality of HCl = 0.1 N

Volume of solution = 100 mL

To calculate normality of solution -

\(M_1 \times V_1 = N_2V_2\)

\(0.1 \times 40 = N_2 \times 20\)

\(N_2 = 0 .2N\)

\(\because\) 1L solution contain number of g equivalent = 0.2

0.1L of solution contains no of g equivalent = 0.1 x 0.2 = 0.02g equivalent

\(\because\) Number of g equivalent = \(\frac{\text{mass}}{\text{equivalent mass}}\)

equivalent mass of \(Na_2CO_3 = \frac{106}2 = 53 g\)

equivalent mass of \(K_2CO_3 = \frac{138}2 = 69 g\)

Let say mass of \(Na_2CO_3 = xg\)

\(\therefore\) mass of \(K_2CO_3 = (1.2 - x)g\)

so, we can write-

\(\frac{1.2 -x}{69} + \frac x{53} = 0.02\)

⇒ \(x = 0.59 g\)

weight of \(Na_2CO_3 \) in mixture is \(= 0.59g\)

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