Question

A solution containing 7g of a solute $(molar\text{ }mass\text{ }210\text{ }g\text{ }mo{{l}^{-1}})$ in 350g of acetone raised the boiling point of acetone from ${{56}^{\circ }}C\text{ }to\text{ }{{56.3}^{\circ }}C$.The value of ebullioscopic constant of acetone in K $kg\text{ }mo{{l}^{-1}}$ $(molar\text{ }mass\text{ }210\text{ }g\text{ }mo{{l}^{-1}})$is :
A. 2.66
B. 3.15
C. 4.12
D. 2.86

Answer 1

To find the solution we will first calculate the molality of solution (m), elevation in boiling point $\Delta {{T}_{b}}$ and then we will find the ebullioscopic constant ${{k}_{b}}$ by using the formula
$\Delta {{T}_{b}}={{k}_{b}}\times m$

Complete Step by step solution:
Initial boiling point of acetone is denoted by ${{T}_{b}}^{\circ }={{56}^{\circ }}C=\left( 273+56 \right)K=329K$
Final boiling point of acetone is denoted by ${{T}_{b}}={{56.3}^{\circ }}C=\left( 273+56.3 \right)K=329.3K$
Mass of solute given is 7g
Mass of solution given is (350+7) g=357g
Firstly, we will calculate the molality of solution (m) as:
$Molality=\dfrac{Number\text{ }of\text{ }moles\text{ }of\text{ }solute}{kg\text{ }of\text{ }solvent}$
$Number\text{ }of\text{ }moles\text{ }of\text{ }solute=\dfrac{mass\text{ }of\text{ }solute}{molar\text{ }mass\text{ }of\text{ }solute}$
Now we can write the equation of molality as:
$Molality=\dfrac{\dfrac{mass\text{ }of\text{ }solute}{molar\text{ }mass\text{ }of\text{ }solute}}{kg\text{ }of\text{ }solvent}$
Substituting the values given in the equation we get:

& m=\dfrac{\dfrac{7}{210}}{\dfrac{350}{1000}} \\\ \implies & \dfrac{10}{105} \\\ \end{aligned}$$ We know that the formula of depression in boiling point is given by: $$\Delta {{T}_{b}}={{k}_{b}}.m$$ Where, ${{k}_{b}}$ =ebullioscopic constant $\Delta {{T}_{b}}$= elevation in boiling point that is equal to ${{T}_{b}}-{{T}_{b}}^{\circ }$ $$\begin{aligned} \implies & \left( 329.3-329 \right) \\\ \implies & 0.3K \\\ \end{aligned}$$ Now, we will find the ebullioscopic constant of acetone ${{k}_{b}}$= $$\begin{aligned} & \Delta {{T}_{b}}={{k}_{b}}\times m \\\ & \therefore 0.3={{k}_{b}}\times \dfrac{10}{105} \\\ & {{k}_{b}}=0.3\times \dfrac{105}{10} \\\ & {{k}_{b}}=3.15K\text{ }Kg\text{ }mo{{l}^{-1}} \\\ \end{aligned}$$ **Hence, we can conclude that the correct option is (b) that is the value of ebullioscopic constant of acetone in K $kg\text{ }mo{{l}^{-1}}$ is 3.15.** **Note:** \- We should not get confused in the terms molality and molarity. As, molality is the ratio of the number of moles of solute in kg of solvent. Molality is denoted by m. \- Whereas, molarity is the ratio of number of moles of solute in volume of solution. Molarity is denoted by M. \- One should not forget to write the unit in the solution.