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Grade 11Physical Chemistry

2.48 g of Na2S2O3.xH2O was dissolved per litre of the solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution. Find out the value of x.

Profile image of Chinmay Shashank Joshi
8 Years agoGrade 11
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1 Answer

Profile image of Chhavi Jain
8 Years ago
Dear Student ,
 
We will use the formula :
  Molarity (M) = no. of moles of solute / volume of solution in Litres
 
Na2S2O3.xH2O was dissolved in 1L of some solution. Now, 20 mL of this solution required 10mL of 0.01M iodine solution.
                               no. of moles = M * V  ( from the above formula )
                               n ( iodine solution)= 0.01 * 10*10-3 = 10-4 moles
                                 
therefore , n ( solution) = 10-4  = Molarity of solution * volume of solution
                         
                                        10-4 = Msol * 20 * 10-3
                                                 Msol = 0.005
 
Now, 2.48 g of Na2S2O3.xH2O  is dissolved in 1L this solution .  So, it is behaving as solute . Again using the same formula we will find the value of x :
 
                                                 0.005 =  moles of solute / 1 L
                                                  0.005 = given mass / molar mass
                                                 
                                                 0.005 = 2.48 / {46+64+48+x (18)}
                                        Now, solve this equation to get the value of x = ?
                         
 
Chemistry Consultant 
Askiitians 

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