An aqueous solution of 4.5 g oxalic acid (H_ | vedclass

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(B) $1$. Calculate the molar mass of oxalic acid ($H_2C_2O_4 \cdot 2H_2O$ is usually implied,but for anhydrous $H_2C_2O_4$,molar mass = $90 \ g/mol$).

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(B) $1$. Calculate the molar mass of oxalic acid ($H_2C_2O_4 \cdot 2H_2O$ is usually implied,but for anhydrous $H_2C_2O_4$,molar mass = $90 \ g/mol$).$2$. Molarity of the solution = $\frac{	ext{mass}}{	ext{molar mass} 	imes 	ext{volume in L}} = \frac{4.5}{90 	imes 0.250} = 0.2 \ M$.$3$. Normality of the oxalic acid solution = $	ext{Molarity} 	imes n	ext{-factor} = 0.2 	imes 2 = 0.4 \ N$.$4$. Using the law of equivalence: $N_1V_1 = N_2V_2$.$5$. $0.4 \ N 	imes 20 \ mL = 0.1 \ N 	imes V_{NaOH}$.$6$. $V_{NaOH} = \frac{0.4 	imes 20}{0.1} = 80 \ mL$.

An aqueous solution of $4.5 \ g$ oxalic acid $(H_2C_2O_4)$ is made up to $250 \ mL$. Calculate the volume of $0.1 \ N$ $NaOH$ required to neutralize completely $20 \ mL$ of this solution.

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