Question

To determine the amount of sodium hydroxide present in the given solution, provided with n/20 standard oxalic acid solution.
Please Give The Observation Tabel Of The Given​

Answer 1

Answer:

Explanation:Calculations

Mass of oxalic acid dissolved in 100ml of standard solution = y g

Strength of oxalic acid = y [latex]\times[/latex] 10 g/L

Normality (N) of standard oxalic acid = Strength/ Eq.wt = [latex]y \times \frac{10}{63.04}[/latex] = N

Normality (N1) of sodium hydroxide solution

N1 [latex]\times[/latex] V1 = N [latex]\times[/latex] V

Therefore,

N1 = [latex]N \times \frac{V}{V_{1}}[/latex]

Normality (N2) of given oxalic acid solution

N2 [latex]\times[/latex] V2 = N1 [latex]\times[/latex] V1

N2 = [latex]N_{1}\times \frac{V_{1}}{V_{2}}[/latex]

Strength of given oxalic acid = N2 x 63.04 g/L