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» <html><body><p></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> mL "of" (N)/(2) NaOH` solution <br/> `=10 mL " of" (N)/(2)<a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a>` solution <br/> `-=50 mL " of" (N)/(10)HCl` solution <br/> Volume of `(N)/(10)HCl` used for neutralisation =300-50=250 mL <br/> 250 mL of `(N)/(10)HCl-=250 mL " of" (N)/(10)` <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> carbonate solution <br/> Let the equivalent mass of metal carbonate be E. <br/> Mass of metal carbonate present in solution <br/> `=(NxxExxV)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)=1.725` <br/> `=(1xxExx250)/(10xx1000)=1.725` <br/> `=(E )/(40)=1.725` <br/> `E=40xx1.725=69`</body></html>