25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was :

25 mL of a solution of barium hydroxide on titration with a 0.1 molar

1.	25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was :
Answer» <html><body><p>0.28<br/>0.35<br/>0.07<br/>0.1</p>Solution :`underset("1 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>")(Ba(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)_(2))+underset("2 mol")(2HCl)rarrBaCl_(2)+2H_(2)O`<br/> No. of moles of `HCl=(("0.1 mol"^(-1)))/(("1L"))xx("0.035L")` <br/> `=3.5xx10^(-3)` mol <br/> No. of moles of `Ba(OH)_(2)=(3.5xx10^(-3))/(2)` <br/> `=1.75xx10^(-3)` mol <br/> <a href="https://interviewquestions.tuteehub.com/tag/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> of `Ba(OH)_(2)` solution <br/> `= ("No. of moles of Ba"("OH")_(2))/("Volume of solution in litres")` <br/> `=((1.75xx10^(-3)"mol"))/(("0.025 L"))=0.07"mol L"^(-1)` <br/> `=0.07M`.</body></html>