7.5 g of an acid are dissolved per litre of the solution. 20 mL of this acid solution required 25 mL of N//15 NaOH solution for complete neutralisation. Calculate the equivalent mass of the acid.

7.5 g of an acid are dissolved per litre of the solution. 20 mL of thi

1.	7.5 g of an acid are dissolved per litre of the solution. 20 mL of this acid solution required 25 mL of N//15 NaOH solution for complete neutralisation. Calculate the equivalent mass of the acid.
Answer» <html><body><p></p>Solution :Calculation of the normality of the acid <br/> Volume of acid taken = 20 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> <br/> volume of base required = <a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> mL <br/> Normality of base `= N//15` <br/> Applying normality equation : `ubrace(N_(1)xxV_(1))_("Acid") equiv ubrace(N_(2) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> V_(2))_("Base")`<br/> `N_(1) xx (20 mL) = ((1)/(15) N) xx 25 mL` <br/> `N_(1) = ((1)/(15) N) xx ((25 mL))/((20 mL)) = (1)/(12) N` <br/> Step <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>. Calculation of the equivalent mass of the acid <br/> Amount of acid dissolved per litre of solution = 7.5 g <br/> `:.` Strength of the acid `= 7.5 g L^(-1)` <br/> Equivalent mass of the acid `= ("Strength of the acid")/("Normality of acid") = ((7.5 g L^(-1)))/((1//12 N)) = 90.0`</body></html>