Answer

Verified

448.5k+ views

**Hint:**First find out the normality concentration of oxalic acid. Then using the formula ${N_1}{V_1} = {N_2}{V_2}$ you can find out the volume of NaOH required for complete neutralisation.

**Complete answer:**

-Normality: $N = \dfrac{{weight}}{{Eq.wt.}} \times \dfrac{{1000}}{{V(mL)}}$ (1)

Where eq.wt. is the equivalent weight.

Complete step by step answer:

-Oxalic acid has a molecular formula of: $HOOC - COOH$ or ${C_2}{H_2}{O_4}$. Oxalic acid dihydrate has a formula of ${C_2}{H_2}{O_4}.2{H_2}O$

Molecular weight of ${C_2}{H_2}{O_4}.2{H_2}O$ is = 90 + 18(2) = 126 g/mole

-To find the volume of NaOH required to neutralise 10 mL oxalic acid solution we will be using the following formula:

${N_1}{V_1} = {N_2}{V_2}$ (2)

Where ${N_1}$, ${V_1}$ are normality and volume for oxalic acid and ${N_2}$, ${V_2}$ are normality and volume for NaOH base.

-So, first we need to find out the normality of oxalic acid.

Since oxalic acid has 2 H atoms that it can lose to show acidic character, it means that n-factor for oxalic acid dihydrate is = 2. Its molecular weight is 126 g/mole.

Equivalent weight of oxalic acid = molecular weight / n-factor

= 126 / 2

= 63

The weight given in the question = 6.3 g

Using equation (1) we will find out the normality of oxalic acid dihydrate:

$N = \dfrac{{weight}}{{Eq.wt.}} \times \dfrac{{1000}}{{V(mL)}}$

= $\dfrac{{6.3}}{{63}} \times \dfrac{{1000}}{{250}}$

= 0.4 N

-Now we will use equation (2) to find out the volume of NaOH required for complete neutralisation.

The question gives the value of: ${N_1}$ = 0.4 N (calculated above), ${V_1}$ = 10 mL and ${N_2}$= 0.1 N. We need to find out the value of ${V_2}$.

${N_1}{V_1} = {N_2}{V_2}$

0.4×10 = 0.1×${V_2}$

${V_2} = \dfrac{{0.4 \times 10}}{{0.1}}$

= 40 mL

So, the volume of 0.1 N NaOH required to neutralise 0.4 N and 10 mL solution of oxalic acid dihydrate is 40 mL.

**So, the correct answer is “Option A”.**

**Note:**The molecular formula of oxalic acid is ${C_2}{H_2}{O_4}$ and its molecular weight is 90 g/mole while that of oxalic acid dihydrate is ${C_2}{H_2}{O_4}.2{H_2}O$ and molecular weight is 126 g/mole. Although both will have the same value of n-factor to be 2, they will have different equivalent weights due to difference in molecular weight. So while calculating the normality in this question use molecular weight to be 126 g/mole.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

How do you graph the function fx 4x class 9 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What organs are located on the left side of your body class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell