Question

10. 1.575 g of oxalic acid (COOH),.xH,O are dissolved
in water and the volume made upto 250 mL. On
titration 16.68 mL of this solution requires 25 mL of
N/15 NaOH solution for complete neutralization.
Calculate x​

Answer 1

Explanation:

Meq of oxalic acid in 16.68ml=Meq of NaOH

⇒25×115

Meq of oxalic acid in 250ml

⇒25×115×2.5016.68

⇒24.98

⇒1.575(90+18x)2×1000=24.98

x≈2(approx)

Hence (c) is the correct answer.

Answer 2

The value of 'x' or number of water molecules in oxalic acid are 2

Explanation:

The relation between normality and molarity is:

$N=n\times M$

where,

N = normality of NaOH = $\frac{1}{15}N$

n = n-factor = 1 (for NaOH)

M = molarity = ?

Putting values in above equation, we get:

$M=\frac{\frac{1}{15}}{1}=\frac{1}{15}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Mass of $C_2H_2O_4.xH_2O$ = 1.575 g

Molar mass of

Volume of solution = 250 mL = 0.250 L      (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$\text{Molarity of }C_2H_2O_4.xH_2O=\frac{1.575}{(90+18x)\times 0.250}$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is oxalic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=\frac{1.575}{(90+18x)\times 0.250}M\\\\V_1=16.68mL\\n_2=1\\M_2=\frac{1}{15}M\\\\V_2=25mL$

Putting values in above equation, we get:

$2\times \frac{1.575}{(90+18x)\times 0.250}\times 16.68=1\times \frac{1}{15}\times 25\\\\x=2$

Learn more about neutralization and molarity of the solution:

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