2.48 g of Na_(2)S_(2)O_(3)xH_(2)O was dissolved per litre of the solution 20mL of tis solution required 10 ml 0.01 m iodin e solution find out the value of x

2.48 g of Na_(2)S_(2)O_(3)xH_(2)O was dissolved per litre of the solut

1.	2.48 g of Na_(2)S_(2)O_(3)xH_(2)O was dissolved per litre of the solution 20mL of tis solution required 10 ml 0.01 m iodin e solution find out the value of x
Answer» <html><body><p><br/></p>Solution :The balanced eqaution for the <a href="https://interviewquestions.tuteehub.com/tag/redox-2246707" style="font-weight:bold;" target="_blank" title="Click to know more about REDOX">REDOX</a> <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> is : `2Na_(2)Cr_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2` <a href="https://interviewquestions.tuteehub.com/tag/nal-569309" style="font-weight:bold;" target="_blank" title="Click to know more about NAL">NAL</a> <br/> Let the molarity of `Na_(2)S_(2)O_(3)xH_(2)O "solution" =M_(1)` <br/><br/> Applying molarity equation to the above redox reaction we have `(M_(1)xx20)/(2)(Na_(2)S_(2)O_(3))=(10xx.01)/(1)(I_(2))` <br/> `therefore M_(1)=0.01 M` <br/> But the actual amount dissolved =2.48g <br/> Equating these <a href="https://interviewquestions.tuteehub.com/tag/values-25920" style="font-weight:bold;" target="_blank" title="Click to know more about VALUES">VALUES</a> we have `(158+18x)xx0.01=2.48 or x=5`</body></html>