Question

a.) The boiling point of benzene is $353.23K$. When $1.80g$ of a non-volatile non-ionization solute was dissolved in $90g$ of benzene, the boiling point raised to $354.11K$
Calculate the molar mass of the solute. $\left[ K{}_{b}\text{ for benzene = 2}\text{.53K kg mo}{{\text{l}}^{-1}} \right]$
b.) Define:
i.) The molality of a solution
ii.) Isotonic solutions.

Answer 1

Hint: We know that the expression for molar mass of a non-volatile non-ionization solute is :
${{M}_{2}}=\dfrac{{{K}_{b}}\times {{w}_{2}}\times 1000}{\Delta {{T}_{b}}\times {{w}_{1}}}$
- Using this expression we can find the required molar mass of the solute. Also try to understand the standard definitions of molality and the isotonic solutions.

Complete Solution :
a.) Given data is as shown below
The boiling point of benzene is $353.23K$
Boiling point is raised to $354.11K$
Therefore the temperature raised is $\Delta T=0.88K$
Also given $K{}_{b}\text{ for benzene = 2}\text{.53K kg mo}{{\text{l}}^{-1}}$
Weight of the benzene is ${{w}_{1}} = 90g$
Weight of a non-volatile non-ionization solute is ${{w}_{2}}=1.80g$
Substitute all the above given values in the formula
${{M}_{2}}=\dfrac{{{K}_{b}}\times {{w}_{2}}\times 1000}{\Delta {{T}_{b}}\times {{w}_{1}}}$
Where the temperature raised is $\Delta T$
$K{}_{b}\text{ }$ is the ionization constant
Weight of the solvent is ${{w}_{1}}$
Weight of a non-volatile non-ionization solute is ${{w}_{2}}$
Molar mass of a non-volatile non-ionization solute is ${{M}_{2}}$

To find the required molar mass of the non-volatile non-ionization solute
$\Rightarrow {{M}_{2}}=\dfrac{2.53\times 1.80\times 1000}{0.88\times 90}$
$\Rightarrow {{M}_{2}}=57.5gmo{{l}^{-1}}$
Therefore the required molar mass of the solute is ${{M}_{2}}=57.5gmo{{l}^{-1}}$

b.)
i.) Molality:
Molality can be defined as the amount or the number of moles of solute dissolved per one kilogram of the solvent.
This can be mathematically represented as
$\text{Molality = }\dfrac{\text{number of moles of solute}}{\text{volume of solvent }\left( \text{in kg} \right)}$
ii.) Isotonic solutions:
Two solutions are said to be isotonic solutions if they have the same osmotic pressure at a given temperature.

Note: When a non-volatile non-ionization solute is mixed in a solvent then the physical properties like boiling point and melting point of the solvent will either be elevated or depressed. We can find the amount of rise of the properties using the above formulas. Concentration of a solution can be expressed in different ways like molarity, molality, normality etc are the terms that show the amount of solute present in the solution.