Answer

Verified

427.5k+ views

**Hint:**Two solutions are called isotonic if they have the same osmotic pressure. To calculate the apparent degree of ionisation of NaCl, you need to first calculate the van’t Hoff factor for NaCl. You should know the formula for calculating osmotic pressure.

**Complete Solution :**

- Given 0.1 M solution of NaCl this means, we are given molarity of NaCl solution = 0.1 M.

NaCl solution is found to be isotonic with 1.1% of urea solution. We know that two solutions are said to be isotonic when they have the same osmotic pressure.

Formula of calculating osmotic pressure, denoted by $\pi $ is:

$\pi = i\dfrac{n}{V}RT = iCRT$

- Where, $i$ is the van’t Hoff factor for the solute, n is the number of moles, V is volume of solution, R is the gas constant, and T is temperature.

- Now, ${\pi _{NaCl}} = i \times 0.1 \times RT$

Given, 1.1% of urea solution. 1.1 g of urea is present in 100 mL of solution. Molar mass of urea, having chemical formula $N{H_2}CON{H_2}$, is 60 g. Thus,

Number of moles of urea = $\dfrac{{1.1}}{{60}}$

Volume of urea solution = 0.1 L

Urea is a nonelectrolyte, therefore van't Hoff factor, $i = 1$ for urea.

So, ${\pi _{Urea}} = i\dfrac{n}{V}RT = \dfrac{{1.1}}{{60 \times 0.1}}RT$

Since, ${\pi _{NaCl}} = {\pi _{Urea}}$

$i \times 0.1 \times RT = \dfrac{{1.1}}{{60 \times 0.1}}RT$

$\therefore i = 1.83$

- Ionization reaction of NaCl is:

\[\begin{align}

& \text{Reaction}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ NaCl\to N{{a}^{+}}+C{{l}^{-}} \\

& \begin{matrix}

\text{Initial}\ \text{moles} & \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ & 0\ \ \ \ & 0 \\

\end{matrix} \\

& \begin{matrix}

\text{Moles}\ \text{at time T}\ \ \ \ & 1-\alpha \ \ \ & \alpha \ \ \ & \alpha \\

\end{matrix} \\

\end{align}\]

Where, $\alpha $ is the degree of ionisation.

- Thus, van’t Hoff factor, $i$ for NaCl:

$i = \dfrac{{{\text{Number}}\;{\text{of}}\;{\text{moles}}\;{\text{after}}\;{\text{ionisation}}}}{{{\text{Number of moles before ionisation}}}} = \dfrac{{1 + \alpha }}{1} = 1 + \alpha $

$1.83 = 1 + \alpha $

$ \Rightarrow \alpha = 0.83$

Hence, the degree of ionisation of NaCl is 0.83.

**Note:**Van’t Hoff factor predicts the extent of dissociation or association of solutes in the solvents. It is defined as:

$i = \dfrac{{{\text{Observed colligative property}}}}{{{\text{Normal colligative property}}}}$

- Osmotic pressure is one of the colligative properties. Also, keep in mind that the van't Hoff factor, $i = 1$ for urea because it is a nonelectrolyte, shows no association or dissociation.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE