Question

The density of a 3 M $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$solution is $1.25\text{ }g\text{ }m{{l}^{-1}}$. Calculate the percentage by mass of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$and molalities of $N{{a}^{+}}$and ${{S}_{2}}{{O}_{3}}^{2-}$ions.

Answer 1

Hint: The mass percentage of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ can be calculated using the mass of the 3M solution of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ and mass of 1000ml of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ solution. Moles of water and moles of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ can give us the mole fraction. The molality of cations and anions formed by dissociation of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ can be calculated using the moles of $N{{a}^{+}}$,${{S}_{2}}{{O}_{3}}^{2-}$ and mass of water (in kg).

Complete step by step solution:
-According to question,
Molarity of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$= 3M
-The density of sodium thiosulfate $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$solution = 1.25 g/mL
We can calculate the mass of 1000mL of sodium thiosulfate solution using the density given to us.
Mass of 1000mL of sodium thiosulfate ($N{{a}_{2}}{{S}_{2}}{{O}_{3}}$) solution:
 $\begin{align}
& Density=\dfrac{Mass}{Volume} \\
& \Rightarrow Mass=Density\times Volume \\
& =1.25\text{ g m}{{\text{L}}^{-1}}\times 1000mL=1250g \\
\end{align}$
-Calculating the mass of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ using the molar mass and molarity of the solution.
As we know,
$Mass=Molarity\times \text{Molar mass}$
Mass of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}=3M\times 158gmo{{l}^{-1}}=474g$
-Using the mass of 1000ml of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$, we can calculate the mass percentage of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$.
The formula to calculate the mass percentage is,
$\text{Mass percentage = }\dfrac{\text{Mass of solute (in g)}}{\text{Mass of solution (in g)}}\times 100\%$
The mass percentage of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$$=\dfrac{474g}{1250g}\times 100\%=37.92\%$
-Molalities of $N{{a}^{+}}$ and ${{S}_{2}}{{O}_{3}}^{2-}$can be calculated using the number of moles and mass of water (in kg).
$1N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to 2N{{a}^{+}}+1{{S}_{2}}{{O}_{3}}^{2-}$
Moles of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}=\dfrac{\text{Given mass(g)}}{\text{Molar mass}}=\dfrac{474}{158g/mol}=3mol$
We shall calculate the mass of water by subtracting the mass of 1000mL of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ solution from the mass of 3M $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$.
Mass of water = Mass of 1000 mL of the solution – Mass of 3M $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$
Mass of water = 1250 – 474g = 776g
The number of moles of sodium cation is calculated as,
Number of moles of $N{{a}^{+}}=2\times \text{ Number of moles of }N{{a}_{2}}{{S}_{2}}{{O}_{3}}=2\times 3=6moles$
Therefore, the molality of sodium ions $=\dfrac{\text{Number of moles of N}{{\text{a}}^{+}}ions}{\text{Mass of water (in kg)}}=\dfrac{6mol}{776g}\times 100=7.73m$
The molarity of ${{S}_{2}}{{O}_{3}}^{2-}=\dfrac{\text{Number of moles of }{{\text{S}}_{2}}{{O}_{3}}^{2-}\text{ions}}{\text{Mass of water (in kg)}}=\dfrac{3mol}{776g}\times 1000=3.865m$

Note: The molality as a measure of concentration depends on the masses of solute and solvent which are affected by changes in temperature and pressure. In contrast, volumetrically prepared solutions such as molar concentration or mass concentration are subjected to change concerning temperature and pressure change. In several applications, this offers a major advantage as the mass of the amount of a substance is often more necessary than its volume. Another advantage of molality is that molality of one solute in a solution is independent of the absence or presence of other solutes in the solution.