1.725 g of a metal carbonate is mixed with 300 mL of (N)/(10)HCl. 10 mL " of" (N)/(2) sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.

1.725 g of a metal carbonate is mixed with 300 mL of (N)/(10)HCl. 10 m

1.	1.725 g of a metal carbonate is mixed with 300 mL of (N)/(10)HCl. 10 mL " of" (N)/(2) sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.
Answer» Solution :`10 mL "of" (N)/(2) NaOH` solution `=10 mL " of" (N)/(2)HCL` solution `-=50 mL " of" (N)/(10)HCl` solution Volume of `(N)/(10)HCl` used for neutralisation =300-50=250 mL 250 mL of `(N)/(10)HCl-=250 mL " of" (N)/(10)` METAL carbonate solution Let the equivalent mass of metal carbonate be E. Mass of metal carbonate present in solution `=(NxxExxV)/(1000)=1.725` `=(1xxExx250)/(10xx1000)=1.725` `=(E )/(40)=1.725` `E=40xx1.725=69`