Question

Calculate the molarity of Na2S2O3 if 41.22 ml is equivalent to 37.84 ml I2 solution, 35.62 ml of this I2 solution is equivalent to 0.2750 g As2O3 (197.84 g/mol)

Answer 1

As2O3 + 2 I2 + 2 H2O ------> As2O5 + 4 HI

1 mole of As2O3 = 2 moles of I2

197.84 grams. = 508 grams

But, given is 0.2750 grams of As2O3.  

Then, grams of iodine = 508*0.2750/197084 = 0.706 grams

Thus, 35.62ml of I2 contains = 0.706 grams I2

Thus, molarity = (wt/mol wt)* 1000/35.62

Molarity of I2 solution =(0.706/254)* 1000/35.62

= 0.078M

2 Na2S2O3 + I2 ------> 2NaI + Na2S4O6

Given volume of Na2S2O3 = V1 41.22ml, n1= 2

I2 Volume V2= 37.84ml , M2= 0.078M , n2= 1

We know ,

M1V1/n1 = M2 V2/n2

M1*41.22= 2*0.078*37.84

M1 = 2* 0.078*37.8441.22

M1= 2*0.0358= 0.0716 M