Question

The density of 3M solution of sodium thiosulphate ($N{a_2}{S_2}{O_3}$) is 1.56g/mL. Calculate
(A).amount of $N{a_2}{S_2}{O_3} in \% \dfrac{w}{w}$
(B).mole fraction of $N{a_2}{S_2}{O_3}$
(C).Molality of $N{a^ + }$and ${S_2}{O_3}^{ - 2}$ions

Answer 1

Molality is defined as the number of moles in one kilogram of a solvent. Mole fraction of a component in a solution is the number of moles of that constituent divided by total moles in solution.

Complete step by step answer:
For our convenience, let the volume of the solution be 1L = 1000mL.
Therefore,
Molarity of a constituent = $\dfrac{{No.\,of\,moles\,of\,the\,same\,constituent}}{{Volume\,of\,solution(in\,L)}}$
3 = $\dfrac{{No.\,of\,moles\,of\,N{a_2}{S_2}{O_3}}}{1}$
No. of moles of $N{a_2}{S_2}{O_3}$ = 3
Now,
Mass of $N{a_2}{S_2}{O_3}$= 3158g = 474g

And,
Density of solution = 1.56g/mL
Density of solution = $\dfrac{{Mass\,of\,solution}}{{Volume\,of\,solution}}$
1.56 = $\dfrac{{Mass\,of\,solution(in\,gms)}}{{1000}}$
Mass of solution = 1560g

Therefore,
Amount of $N{a_2}{S_2}{O_3}$ in $$

Now,
Mass of solute = (1560-474) g
= 1086 g

Molecular weight of solute (water) = 18 g/mole

So, the moles of solute = $\dfrac{{1086}}{{18}}$
= 60.83 moles
Hence, mole fraction of $N{a_2}{S_2}{O_3}$ = $\dfrac{3}{{3 + 60.83}}$
= 0.046
And, Molality of solution = $\dfrac{{Moles\,of\,solution}}{{Weight\,of\,solvent(in\,kg)}}$
Molality of solution = $\dfrac{3}{{1.086}}$
Molality of solution = 2.76m

Now,
$N{a_2}{S_2}{O_3} \to 2N{a^ + } + {S_2}{O_3}^{ - 2}$
Hence, the molality of $N{a^ + }$= 2 Molality of $N{a_2}{S_2}{O_3}$
And the molality of ${S_2}{O_3}^{ - 2}$= Molality of $N{a_2}{S_2}{O_3}$

Therefore,
the molality of $N{a^ + }$= 2 2.76m
= 5.52m
And
the molality of ${S_2}{O_3}^{ - 2}$= 2.76m

Hence, concluding
amount of $N{a_2}{S_2}{O_3}$ in % $\dfrac{w}{w}$= 30.38%
mole fraction of $N{a_2}{S_2}{O_3}$ = 0.046
Molality of $N{a^ + }$and ${S_2}{O_3}^{ - 2}$ ions = 2.76m

Note: A student can confuse between molality and molarity of a solution. Both are concentration terms. Molarity is measured in $\dfrac{{mol}}{L}$while molality is measured in $\dfrac{{mol}}{{kg}}$.