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.

Explanation

`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`