Question

The weight of iodine required to oxidize $500\,ml\,\,N{a_2}{S_2}{O_3}$ solution, is
A.$6.35\,g$
B.$63.5\,g$
C.$127\,g$
D.$254\,g$

Answer 1

The best approach to find the weight of iodine required to oxidize the given amount of $N{a_2}{S_2}{O_3}$ . The solution is to write the reaction of iodine with sodium thiosulphate. We can observe the elements of reactions individually and figure out the reducing and oxidizing agent. Then to calculate the weight we will use some basic concepts of chemistry.

Complete step-by-step answer: First, we will write a chemical reaction in which sodium thiosulphate is reacting with iodine to form sodium tetrathionate and sodium iodide. The balanced reaction can be written as given below.
$2N{a_2}{S_2}{O_3}\, + \,{I_2}\, \to \,N{a_2}{S_4}{O_6}\, + \,2NaI$
Now we will use the basic concepts of chemistry in while using molar mass we see that sodium thiosulphate reacts with iodine. According to the equation sodium thiosulphate reacts with one mole of iodine, as molar mass of iodine is $253.8\,g$ thus sodium thiosulphate get oxidised with $253.8\,g$ of iodine.
$2N{a_2}{S_2}{O_3}\, + \,{I_2}\, \to \,N{a_2}{S_4}{O_6}\, + \,2NaI$
$2\,moles\,(2N{a_2}{S_2}{O_3}\,)\,,\,\,1\,mole\,({I_2})$
So, according to the above equation, we can conclude that $22.4L$ of $N{a_2}{S_2}{O_3}$ is oxidized by $253.8\,g$ iodine. We know that $1$ mole of iodine contains $253.8\,g$ iodine.
$1\,mole\,of\,{I_2} = \,253.8\,g\,iodine$
$22.4L\,of\,N{a_2}{S_2}{O_3}\,oxidised\,by\, = 253.8\,g\,iodine$
$(22.4ml\, \times 1000)\,of\,N{a_2}{S_2}{O_3}\,oxidised\,by = \,(253.8\,g\,)\,iodine$
Now we need to calculate the weight of iodine required to oxidize $500\,ml\,\,N{a_2}{S_2}{O_3}$ the solution. So as above we convert $22.4L$ into $(ml)$ by multiplying it with $1000$ . So, $224000\,ml\,$ of $N{a_2}{S_2}{O_3}$ is oxidized by $253.8\,g$ iodine. So, for $1ml$ it was,
$1\,ml\,of\,N{a_2}{S_2}{O_3}\,oxidised\,by = \,\dfrac{{(253.8\,g\,)}}{{(22.4\, \times 1000)}}\,iodine$ = $0.011330\,gm$
Now for $500\,ml\,\,N{a_2}{S_2}{O_3}$ we will multiply the mass with $500\,$ thus after solving the equation we get,
$500\,ml\,of\,N{a_2}{S_2}{O_3}\,oxidised\,by = \,(0.01133 \times \,500)\,g\,iodine$
$500\,ml\,of\,N{a_2}{S_2}{O_3}\,oxidised\,by = \,5.89g\,iodine$
It will be near to $6.5\,gm$ .

Therefore, the correct option is (A).

Note: The related questions can be solved using basic relationships. The data can be given in terms of mass, molecules, and moles. The given data can be solved by converting moles into mass and if we multiply the mass with the Avogadro number we will get the number of particles. Thus in the above question we use the moles required for reaction and then change them in mass.