Question

The density of 3M sodium thiosulfate solution $\left( {N{a_2}{S_2}{O_3}} \right)$ is $1.25\,g/mL$. Calculate (i) the percentage of mass of sodium thiosulfate, (ii) the mole fraction of sodium thiosulfate and (iii) molalities of $N{a^ + }$ and ${S_2}{O_3}^{2 - }$ ions.

Answer 1

Hint:We can calculate the mass percentage of $N{a_2}{S_2}{O_3}$ using the mass of $N{a_2}{S_2}{O_3}$ of 3M solution and mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution. We can calculate the mole fraction using the moles of water and the moles of $N{a_2}{S_2}{O_3}$. We can calculate the molality of $N{a^ + }$ and ${S_2}{O_3}^{2 - }$ ions using the moles of $N{a^ + }$ and ${S_2}{O_3}^{2 - }$ and mass of water (in kg).

Complete step by step answer:

Given,
Molarity of sodium thiosulfate solution is 3M.
The density of sodium thiosulfate solution is $1.25\,g/mL$.
(i)
We can calculate the mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution using density of the solution as,
Mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution=$1.25\,\dfrac{g}{{mL}} \times \dfrac{{1000\,mL}}{{1\,}}$
Mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution=$1250\,g$
Mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution is $1250\,g$.
Let us now calculate the mass of $N{a_2}{S_2}{O_3}$ of 3M solution using the molar mass and molarity of the solution.
We know that the molar mass of $N{a_2}{S_2}{O_3}$ is $158\,g\,mo{l^{ - 1}}$.
The mass of $N{a_2}{S_2}{O_3}$ of 3M solution is calculated as,
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$Molarity \times Molar\,mass\,of\,N{a_2}{S_2}{O_3}$
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$3\,M \times 158\,\dfrac{g}{{mol}}$
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$474\,g$
The mass of $N{a_2}{S_2}{O_3}$ of 3M solution is $474\,g$.
Finally, we can calculate the mass percentage of $N{a_2}{S_2}{O_3}$ using the mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution and mass of $N{a_2}{S_2}{O_3}$ of 3M solution.
The formula to calculate the mass percentage is,
Mass percentage=$\dfrac{{Mass\,of\,solute\left( {in\,g} \right)}}{{Mass\,of\,solution\left( {in\,g} \right)}} \times 100\% $
The mass percentage of $N{a_2}{S_2}{O_3}$ is calculated as,
The mass of $N{a_2}{S_2}{O_3}$ of 3M solution is $474\,g$.
Mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution is $1250\,g$.
Mass percentage=$\dfrac{{474\,g}}{{1250g}} \times 100\% $
Mass percentage=$37.92\% $
The mass percentage of $N{a_2}{S_2}{O_3}$ is $37.92\% $.
(ii)
Let us calculate the mole fraction of $N{a_2}{S_2}{O_3}$
In order to find the mole fraction, we should know the moles of $N{a_2}{S_2}{O_3}$ and moles of water. We can find the moles using their weight and molar masses.
We know that the molar mass of $N{a_2}{S_2}{O_3}$ is $158\,g\,mo{l^{ - 1}}$.
The mass of $N{a_2}{S_2}{O_3}$ of 3M solution is calculated as,
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$Molarity \times Molar\,mass\,of\,N{a_2}{S_2}{O_3}$
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$3\,M \times 158\,\dfrac{g}{{mol}}$
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$474\,g$
The mass of $N{a_2}{S_2}{O_3}$ of 3M solution is $474\,g$.
From the calculated mass, let us now calculate the moles of $N{a_2}{S_2}{O_3}$.
Moles of $N{a_2}{S_2}{O_3}$=$\dfrac{{Mass\left( {in\,g} \right)}}{{Molar\,mass}}$
Moles of $N{a_2}{S_2}{O_3}$=$\dfrac{{474\,g}}{{158\,g/mol}}$
Moles of $N{a_2}{S_2}{O_3}$=$3\,mol$
The moles of $N{a_2}{S_2}{O_3}$ is $3\,mol.$
We can calculate the mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution using density of the solution as,
Mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution=$1.25\,\dfrac{g}{{mL}} \times \dfrac{{1000\,mL}}{{1\,}}$
Mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution=$1250\,g$
Mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution is $1250\,g$.
Now, we shall calculate the mass of water by subtracting the mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution to the mass of $N{a_2}{S_2}{O_3}$.
Mass of water=\[{\text{Mass of 1000mL of solution - Mass}}\,\,{\text{of}}\,{\text{3M}}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}\]
Mass of water=\[{\text{1250}}\,{\text{g - 474}}\,{\text{g}}\]
Mass of water=${\text{776}}\,{\text{g}}$
The mass of water is ${\text{776}}\,{\text{g}}$.
From this we shall calculate the moles of water as,
The molar mass of water is $18\,g/mol$.
Moles of water=$\dfrac{{Mass\left( {in\,g} \right)}}{{Molar\,mass}}$
Moles of water=$\dfrac{{776\,g}}{{18\,g/mol}}$
Moles of water=$43.1\,\,mol$
The moles of water is $43.1\,\,mol$.
The mole fraction of $N{a_2}{S_2}{O_3}$ is calculated using the moles of $N{a_2}{S_2}{O_3}$ and the total moles.
The formula to calculate mole fraction of $N{a_2}{S_2}{O_3}$ is,
Mole fraction of $N{a_2}{S_2}{O_3}$=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}}}{{{\text{Moles}}\,{\text{of}}\,{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{ + Moles}}\,{\text{of}}\,{{\text{H}}_{\text{2}}}{\text{O}}}}$
Mole fraction of $N{a_2}{S_2}{O_3}$=$\dfrac{{3\,mol}}{{3\,mol + 43.1\,mol}}$
Mole fraction of $N{a_2}{S_2}{O_3}$=$\dfrac{{3\,mol}}{{46.1\,mol}}$
Mole fraction of $N{a_2}{S_2}{O_3}$=${\text{0}}{\text{.065}}$
The mole fraction of $N{a_2}{S_2}{O_3}$ is ${\text{0}}{\text{.065}}$.
(iii)
We can find the molalities of $N{a^ + }$ and ${S_2}{O_3}^{2 - }$ ions using the number of moles and mass of water (in kg).
$1\,N{a_2}{S_2}{O_3}\xrightarrow{{}}2N{a^ + } + 1{S_2}{O_3}^{2 - }$
The mass of $N{a_2}{S_2}{O_3}$ of 3M solution is calculated as,
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$Molarity \times Molar\,mass\,of\,N{a_2}{S_2}{O_3}$
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$3\,M \times 158\,\dfrac{g}{{mol}}$
Mass of $N{a_2}{S_2}{O_3}$ of 3M solution=$474\,g$
The mass of $N{a_2}{S_2}{O_3}$ of 3M solution is $474\,g$.
From the calculated mass, let us now calculate the moles of $N{a_2}{S_2}{O_3}$.
Moles of $N{a_2}{S_2}{O_3}$=$\dfrac{{Mass\left( {in\,g} \right)}}{{Molar\,mass}}$
Moles of $N{a_2}{S_2}{O_3}$=$\dfrac{{474\,g}}{{158\,g/mol}}$
Moles of $N{a_2}{S_2}{O_3}$=$3\,mol$
The moles of $N{a_2}{S_2}{O_3}$ is $3\,mol.$
Now, we shall calculate the mass of water by subtracting the mass of 1000mL of $N{a_2}{S_2}{O_3}$ solution to the mass of 3M $N{a_2}{S_2}{O_3}$.
Mass of water=\[{\text{Mass of 1000mL of solution - Mass}}\,\,{\text{of}}\,{\text{3M}}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}\]
Mass of water=\[{\text{1250}}\,{\text{g - 474}}\,{\text{g}}\]
Mass of water=${\text{776}}\,{\text{g}}$
The mass of water is ${\text{776}}\,{\text{g}}$.
The number of moles of $N{a^ + }$ is calculated as,
Number of moles of $N{a^ + }$ ion=${\text{2 X No}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
Number of moles of $N{a^ + }$ ion=${\text{2 X 3}}$
Number of moles of $N{a^ + }$ ion=${\text{6}}$
Therefore, we can calculate the molality of sodium ions as,
Molality of sodium ions=$\dfrac{{No\,of\,moles\,of\,N{a^ + }\,ions}}{{Mass\,of\,water\left( {in\,kg} \right)}}$
Molality of sodium ions=$\dfrac{{6\,mol}}{{776\,g}} \times 1000$
Molality of sodium ions=$7.73\,m$
The molality of sodium ions is $7.73\,m$.
The molality of ${S_2}{O_3}^{2 - }$ will be calculated as,
Molality of ${S_2}{O_3}^{2 - }$=$\dfrac{{No\,of\,moles\,of\,{S_2}{O_3}^{2 - }\,ions}}{{Mass\,of\,water\left( {in\,kg} \right)}}$
Molality of ${S_2}{O_3}^{2 - }$=$\dfrac{{3\,mol}}{{776\,g}} \times 1000$
Molality of ${S_2}{O_3}^{2 - }$=$3.865\,m$
The molality of ${S_2}{O_3}^{2 - }$ is $3.865\,m$.

Note:
We have to know the primary advantage of using molality as a measure of concentration is that molality depends on the masses of solute and solvent that are unaffected by changes in temperature and pressure.
In contrast, solutions that are prepared volumetrically such as molar concentration or mass concentration are subjected to change with respect to temperature and pressure change. In several applications, this is a major advantage because the mass, or the amount, of a substance is often more necessary than its volume.
Another advantage of molality is that the molality of one solute in a solution is independent of the absence (or) presence of other solutes.