Question

(a) Define the following terms: (i) Molarity (ii) Molal elevation constant $\left( {{k}_{b}} \right)$
(b) A solution containing 15g urea (molar mass$\text{=60gmo}{{\text{l}}^{\text{-1}}}$) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass $\text{=190gmo}{{\text{l}}^{\text{-1}}}$) in water. Calculate the mass of glucose present in one litre of its solution.

Answer 1

Hint: Brush up the concept of solutions and colligative properties. Start by calculating the number of moles of urea. Using Van’t Hoff’s equation $\pi V=nRT$, calculate the osmotic pressure of urea solution. Then use that value and calculate molar mass of glucose by using the same equation.

Complete step by step answer:
- Molarity is the number of moles of solute present in 1 litre volume of the solution. It is expressed in mol/L. It is a volume dependent quantity and volume changes with change in temperature.
- Molal elevation constant $\left( {{k}_{b}} \right)$ is defined as the elevation of boiling point produced, when one mole of solute is dissolved in 1kg of solvent. It is also known as ebullioscopic constant.
- Now, let’s solve the problem. We need to find the molar mass of glucose. It is given that glucose solution is isotonic with urea solution. So, let’s first find the number of moles of urea.

- We know that, $Number\,of\,moles=\dfrac{weight}{molar\,mass}=\dfrac{W}{M}$
- For urea, number of moles, ${{n}_{1}}=\dfrac{15}{60}=0.25$

- Using Van’t Hoff’s general solution equation, $\pi V = nRT$, let’s calculate the osmotic pressure of urea.
$\begin{align}
  & \pi V=nRT \\
 & \pi =\dfrac{{{n}_{1}}RT}{V} = \dfrac{0.25\times 0.0821\times 298}{1} = 6.116atm \\
\end{align}$

- Now, we know, the osmotic pressure of urea solution is 6.116atm which will be the same for glucose solution since they are isotonic to each other.

- Therefore, using the osmotic pressure value, let's find the number of moles of glucose.
$\begin{align}
  & \pi V={{n}_{2}}RT \\
 & {{n}_{2}}=\dfrac{\pi V}{RT}=\dfrac{6.116\times 1}{0.0821\times 298}=0.25mol \\
\end{align}$
- Therefore, the number of moles of glucose present is 0.25mol.
- Now, mass of glucose, $W=M\times {{n}_{2}}=190\times 0.25 = 47.5g$
- Therefore, the mass of glucose is 47.5g in one litre.

Note: Remember, in isotonic solutions, the osmotic pressure of solutions is the same. Mass of a solute is the product of the number of moles of that solute present in the solution and molar mass of that solute. Remember colligative properties depend only on the number of solute particles present in the solution and not the nature of solute particles.