Question

a.Explain the following
i)Henry’s law about dissolution of a gas in a liquid.
ii)Boiling point elevation constant for a solvent.
b.A solution of glycerol ${{C}_{3}}{{H}_{8}}{{O}_{2}}$ in water was prepared by dissolving some glycerol in $500g$ of water. This solution has a boiling point of ${{100.42}^{o}}C$ . What mass of glycerol was dissolved to make this solution? (${{K}_{b}}$ for water \[=0.512\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}\])

Answer 1

Hint: In this problem, we first briefly discuss the statement of Henry’s law and we write about Henry's law about dissolution of a gas in a liquid. Then we find the boiling point elevation constant for a solvent. Finally we find the mass of glycerol which was dissolved in $500g$of water by using the formula $\Delta {{T}_{b}}=\dfrac{{{K}_{b}}\times {{W}_{B}}\times 1000}{{{M}_{b}}\times {{W}_{A}}}$ .

Complete step by step answer:
(a)
(i) Henry’s law is a gas law which states that the amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid when the temperature is kept constant. The constant of proportionality for this relationship is known as Henry’s law constant usually written by ‘k ‘. The mathematical formula of Henry’s law is given by: \[P\text{ }=\text{ }k\text{ }.C\] Where, ‘P’ denotes the partial pressure of the gas in the atmosphere above the liquid. ‘C’ denotes the concentration of the dissolved gas. ‘k’ is the Henry’s law constant of the gas. It can be noted that Henry's law constant can be expressed in two different ways. If the constant is denoted in terms of solubility/pressure, it is referred to as the Henry’s law solubility constant (denoted by ‘H’). On the other hand, if the proportionality constant is denoted in terms of pressure/solubility, it is called the Henry’s law volatility constant (denoted by ‘k ’).
(ii) The boiling point elevation constant or molal elevation constant is a constant quantity for a solute which is related to molar mass and elevation in boiling point by the following expression.
${{K}_{b}}=\dfrac{\Delta {{T}_{b}}\times {{M}_{B}}\times {{W}_{A}}}{{{W}_{B}}\times 1000}$
Where, ${{K}_{b}}$ is the boiling point elevation constant
${{M}_{B}}$ is the molar mass of the solute
${{W}_{A}}$ is the weight of solvent
${{W}_{B}}$ is the weight of solute
$\Delta {{T}_{b}}$ is the elevation in boiling point.

(b) We were given that a solution of glycerol ${{C}_{3}}{{H}_{8}}{{O}_{2}}$ in water was prepared by dissolving some glycerol in $500g$ of water and this solution has a boiling point of ${{100.42}^{o}}C$ .
We were asked to find the mass of glycerol was dissolved to make this solution if the value of ${{K}_{b}}$ for water \[=0.512\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}\].
By using the formula $\Delta {{T}_{b}}=\dfrac{{{K}_{b}}\times {{W}_{B}}\times 1000}{{{M}_{b}}\times {{W}_{A}}}$, we get,
$\begin{align}
& \Delta {{T}_{b}}=\dfrac{{{K}_{b}}\times {{W}_{B}}\times 1000}{{{M}_{b}}\times {{W}_{A}}} \\
& \Rightarrow 0.42=\dfrac{0.512\times {{W}_{B}}\times 1000}{\left[ 3\left( 12 \right)+8\left( 1 \right)+3\left( 16 \right) \right]\times 500} \\
& \Rightarrow {{W}_{B}}=\dfrac{0.42\times 92\times 500}{0.512\times 1000}=37.33g \\
\end{align}$
Therefore, $37.33g$ mass of glycerol was dissolved to make this solution.

Note:
The possibilities of making mistakes in this type of problems are,
It is chance to make a mistake by considering the formula of the elevation of boiling point as $\Delta {{T}_{b}}=\dfrac{{{K}_{b}}\times {{W}_{A}}\times 1000}{{{M}_{b}}\times {{W}_{B}}}$ instead of considering $\Delta {{T}_{b}}=\dfrac{{{K}_{b}}\times {{W}_{B}}\times 1000}{{{M}_{b}}\times {{W}_{A}}}$.
The units of the provided quantities and given weight, molecular weight and others should be the same to calculate the correct answer.