1.575 g of oxalic acid ( COOH )2 x H 2 O are dissolved in water and the volume made upto 250 ml. On titration 8.34 ml of this solution requires 12.5 ml of N / 15 NaOH solution, for complete neutralization. Calculate x.

Question

1.575 g of oxalic acid ( COOH )2 x H 2 O are dissolved in water and the volume made upto 250 ml. On titration 8.34 ml of this solution requires 12.5 ml of N / 15 NaOH solution, for complete neutralization. Calculate x. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Meq. of oxalic acid in 8.34 ml = meq of NaOH =12.5 × (1/15) ∴ Meq of oxalic acid in 250 ml =12.5 × (1/15) × (250/8.34)=24.98 ∴ (1.575/90+(18 x)2) × 1000=24.98, x=2

Q. $1.575 g$ of oxalic acid $(COO H)_{2} x H_{2} O$ are dissolved in water and the volume made upto $250 m l$ . On titration $8.34 m l$ of this solution requires $12.5 m l$ of $N /15 N a O H$ solution, for complete neutralization. Calculate $x$ .

Meq. of oxalic acid in $8.34 m l =$ meq of $N a O H$
$= 12.5 \times \frac{1}{15}$
$∴$ Meq of oxalic acid in $250 m l$
$= 12.5 \times \frac{1}{15} \times \frac{250}{8.34} = 24.98$
$∴ \frac{1.575}{90 + \frac{18 x}{2}} \times 1000 = 24.98, x = 2$

Q. 1.575g of oxalic acid (COOH)2​xH2​O are dissolved in water and the volume made upto 250ml. On titration 8.34ml of this solution requires 12.5ml of N/15NaOH solution, for complete neutralization. Calculate x.

Meq. of oxalic acid in 8.34ml= meq of NaOH =12.5×151​ ∴ Meq of oxalic acid in 250ml =12.5×151​×8.34250​=24.98 ∴90+218x​1.575​×1000=24.98,x=2

Q. $1.575 g$ of oxalic acid $(COO H)_{2} x H_{2} O$ are dissolved in water and the volume made upto $250 m l$ . On titration $8.34 m l$ of this solution requires $12.5 m l$ of $N /15 N a O H$ solution, for complete neutralization. Calculate $x$ .

Meq. of oxalic acid in $8.34 m l =$ meq of $N a O H$
$= 12.5 \times \frac{1}{15}$
$∴$ Meq of oxalic acid in $250 m l$
$= 12.5 \times \frac{1}{15} \times \frac{250}{8.34} = 24.98$
$∴ \frac{1.575}{90 + \frac{18 x}{2}} \times 1000 = 24.98, x = 2$