5.3 g of M2 CO 3 is dissolved in 150 mL of 1 N HCl Unused acid required 100 mL of 0.5 N NaOH.Hence, equivalent weight of M is

Question

5.3 g of M2 CO 3 is dissolved in 150 mL of 1 N HCl Unused acid required 100 mL of 0.5 N NaOH.Hence, equivalent weight of M is (A) 23 (B) 12 (C) 24 (D) 13

Tardigrade Admin · Accepted Answer

V mL of 1 N unused HCl =100 mL of 0.5 N NaOH V=50 mL Used 1 N HCl =100 mL 100 mL of 1 N HCl =0.1 equivalent HCl =0.1 equivalent HCl =0.11 equivalent of M2 CO 3 ∴ equivalent mass of M2 CO 3=(2 M+60/2)=(M+30) ∴ (5.3/M+30)=0.1 ∴ M=23

Q. 5.3g of M2​CO3​ is dissolved in 150mL of 1NHCl Unused acid required 100mL of 0.5NNaOH.Hence, equivalent weight of M is

23

12

24

13

VmL of 1N unused HCl=100mL of 0.5NNaOH V=50mL Used 1NHCl=100mL 100mL of 1NHCl=0.1 equivalent HCl =0.1 equivalent HCl =0.11 equivalent of M2​CO3​ ∴ equivalent mass of M2​CO3​=22M+60​=(M+30) ∴M+305.3​=0.1 ∴M=23

Q. $5.3 g$ of $M_{2} C O_{3}$ is dissolved in $150 m L$ of $1 N H Cl$ Unused acid required $100 m L$ of $0.5 N N a O H$ .Hence, equivalent weight of $M$ is