Question

The molarity of ${H_2}S{O_4}$ is $0.8{\text{ M}}$ and its density is $1.06{\text{ g c}}{{\text{m}}^{ - 3}}$ . What will be the concentration of the solution in terms of molality and mole fraction. Calculate the molarity of water if its density is $1000{\text{ kg }}{{\text{m}}^{ - 3}}$ .

Answer 1

Hint : We can find out the mole fraction and molality of the solution by finding out the masses of the ${H_2}S{O_4}$ present in the solution and the mass of solvent. The mass can be obtained by using the given densities of ${H_2}S{O_4}$ and water. With the help of molarity we can find the mass of ${H_2}S{O_4}$ .
$1.$ Number of moles $= {\text{ }}\dfrac{{given{\text{ mass of solute}}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}solute}}$
$2.$ Density $= {\text{ }}\dfrac{{given{\text{ mass }}}}{{volume{\text{ of solution}}}}$
$3.$ Mole Fraction of solute $= {\text{ }}\dfrac{{moles{\text{ of solute}}}}{{moles{\text{ of solute + moles of solvent}}}}$
$4.$ Molality of solution $= {\text{ }}\dfrac{{moles{\text{ of soute}}}}{{mass{\text{ of water}}}}$

Complete Step By Step Answer:
Firstly we will find out the amount of mass of ${H_2}S{O_4}$ present in the solution. Since the molarity of ${H_2}S{O_4}$ is $0.8{\text{ M}}$ . Molarity is the ratio of moles and volume of solution. Therefore it can be written as $0.8{\text{ mol }}{{\text{L}}^{ - 1}}$ . Assuming the volume be unit litre. The moles of ${H_2}S{O_4}$ are now $0.8{\text{ mol }}$ .
We know that,
Number of moles $= {\text{ }}\dfrac{{given{\text{ mass of solute}}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}solute}}$
Molar mass of ${H_2}S{O_4}$ can be calculated as: $\left( {1{\text{ }} \times {\text{ 2}}} \right){\text{ + 32 + }}\left( {4{\text{ }} \times {\text{16}}} \right){\text{ g = 98 g}}$
Now,
mass of ${H_2}S{O_4}$ $=$ moles of ${H_2}S{O_4}$ $\times$ molar mass of ${H_2}S{O_4}$
$= {\text{ 0}}{\text{.8 }} \times {\text{ 98 g}}$
$= {\text{ 78}}{\text{.4 g}}$
Now we will find out the mass of solution:
We know that,
Density $= {\text{ }}\dfrac{{given{\text{ mass }}}}{{volume{\text{ of solution}}}}$
Density is $1.06{\text{ g c}}{{\text{m}}^{ - 3}}$ and the volume of solution is $= {\text{ 1000 c}}{{\text{m}}^3}$
Therefore the mass can be calculated as,
Mass of the solution $=$ density of solution $\times$ volume of solution
$= {\text{ 1}}{\text{.06 }} \times {\text{ 1000 g}}$
$= {\text{ 1060 g}}$
Since the solution comprises both ${H_2}S{O_4}$ and the water. We will now find the mass of water which is,
Mass of water $=$ Mass of solution $-$ Mass of ${H_2}S{O_4}$
$= {\text{ 1060 - 78}}{\text{.4 g}}$
$= {\text{ 981}}{\text{.6 g}}$
Molality of solution $= {\text{ }}\dfrac{{moles{\text{ of soute}}}}{{mass{\text{ of water}}}}$
$= {\text{ }}\dfrac{{0.8}}{{0.981}}{\text{ mol k}}{{\text{g}}^{ - 1}}$
$= {\text{ 0}}{\text{.815 mol k}}{{\text{g}}^{ - 1}}$
We know that,
Mole fraction of ${H_2}S{O_4}$ $= {\text{ }}\dfrac{{moles{\text{ of solute}}}}{{moles{\text{ of solute + moles of solvent}}}}$
Here the solvent is water, so we have to replace it with moles of water.
Mole fraction of ${H_2}S{O_4}$ $= {\text{ }}\dfrac{{0.8{\text{ mol}}}}{{0.8{\text{ mol + }}\dfrac{{98.6}}{{18}}mol}}$
Mole fraction of ${H_2}S{O_4}$ $= {\text{ }}\dfrac{{0.8{\text{ mol}}}}{{0.8{\text{ mol + 54}}{\text{.5 }}mol}}$
Mole fraction of ${H_2}S{O_4}$ $= {\text{ 0}}{\text{.014}}$
Also if density of water is $1000{\text{ kg }}{{\text{m}}^{ - 3}}$ then its molarity is given by,
Molarity $= {\text{ }}\dfrac{{1000{\text{ }}}}{{18}}{\text{M}}$
$= {\text{ 55}}{\text{.5 M}}$

Note :
The mole fraction is calculated specifically for solute and solvent. Here we calculate it for the solute. Mole fractions do not have any units because it is a ratio of two same quantities. For finding molality we do not take care of the volume of the solution. We can take volume as unity when we have to calculate moles from molarity.