25 cm3 of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

Question

25 cm3 of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is (A) 0.045 (B) 0.032 (C) 0.064 (D) 0.015

Tardigrade Admin · Accepted Answer

Moles of oxalic acid = Moles of NaOH V × N =(W/M) (25/1000) × N =(0.064/40) N =(0.064 × 1000/25 × 40)=0.064 ∵ Molarity of oxalic acid = Basicity × Normality =2 × 0.064 =0.032

Q. $25 c m^{3}$ of oxalic acid completely neutralised $0.064 g$ of sodium hydroxide. Molarity of the oxalic acid solution is

0.045

0.032

0.064

0.015

Moles of oxalic acid = Moles of $N a O H$

$V \times N = \frac{W}{M}$

$\frac{25}{1000} \times N = \frac{0.064}{40}$

$N = \frac{0.064 \times 1000}{25 \times 40} = 0.064$

$∵$ Molarity of oxalic acid $=$ Basicity $\times$ Normality

$= 2 \times 0.064$

$= 0.032$

Q. 25cm3 of oxalic acid completely neutralised 0.064g of sodium hydroxide. Molarity of the oxalic acid solution is