25 mL of 0.08 N Mohr's salt solution is oxidised by 20 mL of K2Cr2O7 in acidic medium. The mass of Mohr's salt present in 500 cc is

Question

25 mL of 0.08 N Mohr's salt solution is oxidised by 20 mL of K2Cr2O7 in acidic medium. The mass of Mohr's salt present in 500 cc is (A) 3.96 g (B) 15.68 g (C) 39.6 g (D) 39.2 g

Tardigrade Admin · Accepted Answer

0.08 N of Mohr's salt= 0.08 M
No. of moles in 25 mL of 0.08 M Mohr's salt solution

= 0.08 \times \frac{25}{1000} = 0.002

No. of moles in 500 cc of Mohr's salt solution

0.002 \times \frac{500}{25} = 0.04

Hence, mass of Mohr's salt present in 500 cc

= 0.04 \times 392 = 15.68 g

Q. 25 mL of 0.08 N Mohr's salt solution is oxidised by 20 mL of K2​Cr2​O7​ in acidic medium. The mass of Mohr's salt present in 500 cc is

3.96 g

15.68 g

39.6 g

39.2 g

0.08 N of Mohr's salt= 0.08 M No. of moles in 25 mL of 0.08 M Mohr's salt solution =0.08×100025​=0.002 No. of moles in 500 cc of Mohr's salt solution 0.002×25500​=0.04 Hence, mass of Mohr's salt present in 500 cc =0.04×392=15.68g

Q. 25 mL of 0.08 N Mohr's salt solution is oxidised by 20 mL of $K_{2} C r_{2} O_{7}$ in acidic medium. The mass of Mohr's salt present in 500 cc is

0.08 N of Mohr's salt= 0.08 M
No. of moles in 25 mL of 0.08 M Mohr's salt solution
$= 0.08 \times \frac{25}{1000} = 0.002$
No. of moles in 500 cc of Mohr's salt solution
$0.002 \times \frac{500}{25} = 0.04$
Hence, mass of Mohr's salt present in 500 cc
$= 0.04 \times 392 = 15.68 g$