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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.
one year ago

Chhavi Jain
92 Points

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
one year ago
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• Mind Map
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