If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
Show Hint
Correct Answer: 1.6
Approach Solution - 1
To find the van’t Hoff factor (\(i\)) for the compound \(A_2B\) which is 30% ionised, we need to understand how ionisation impacts \(i\).
Assume initially we have 1 mole of \(A_2B\). Upon ionisation, it dissociates as follows:
\(A_2B \rightarrow 2A^+ + B^-\)
This means, for complete ionisation, from 1 mole of \(A_2B\), we would get 2 moles of \(A^+\) and 1 mole of \(B^-\), equating to 3 moles of ions in total.
However, it is only 30% ionised. Therefore:
- Moles of \(A_2B\) ionised = 0.3 moles
- Moles of \(A_2B\) undissociated = 0.7 moles
From 0.3 moles of \(A_2B\) ionised, ions produced are:
- Moles of \(A^+\) = 2 × 0.3 = 0.6 moles
- Moles of \(B^-\) = 0.3 moles
Total moles after ionisation = moles undissociated + moles of ions = 0.7 + 0.6 + 0.3 = 1.6 moles
The van’t Hoff factor (\(i\)) is defined as:
\(i = \frac{\text{total moles after ionisation}}{\text{initial moles of solute}} = \frac{1.6}{1} = 1.6\)
Approach Solution -2
For the compound \( A_2B \), dissociation in aqueous solution can be represented as:
\[ A_2B \rightleftharpoons 2A^+ + B^{2-} \]
If the degree of ionization (α) = 30% = 0.3, then out of 1 mole of \( A_2B \), 0.3 moles dissociate.
Step 2: Calculate total number of moles after dissociation.
Initial moles of \( A_2B \) = 1
Moles dissociated = 0.3
So, moles of undissociated \( A_2B \) = \( 1 - 0.3 = 0.7 \)
From 0.3 moles of dissociated \( A_2B \):
- Moles of \( A^+ \) formed = \( 2 \times 0.3 = 0.6 \)
- Moles of \( B^{2-} \) formed = 0.3
Total moles after dissociation = \( 0.7 + 0.6 + 0.3 = 1.6 \)
Step 3: Apply the van’t Hoff factor formula.
\[ i = \frac{\text{Total number of moles after dissociation}}{\text{Initial moles}} \] \[ i = 1.6 \]
Step 4: Final Result.
The van’t Hoff factor for 30% ionization of \( A_2B \) is:
\[ \boxed{i = 1.6} \]
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Colligative Properties Questions
- What happens to freezing point of benzene when small quantity of napthalene is added to benzene?
- We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01M & 0.001 M, respectively. The value of vant’ Haft factor (i) for these solutions will be in the order.
- An artificial cell is made by encapsulating \(0.2 \, \text{M}\) glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a \(0.05 \, \text{M}\) solution of NaCl at 300 K is ______ \( \times 10^{-1} \, \text{bar} \). (Nearest Integer)
Given: \(R = 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}\)
Assume complete dissociation of NaCl. - 0.05M CuSO$_4$ when treated with 0.01M K$_2$Cr$_2$O$_7$ gives green colour solution of Cu$_2$Cr$_2$O$_7$. The \[ \text{[SPM : Semi Permeable Membrane]} \]
\(K_2Cr_2O_7\) \(CuSO_4\) Side X SPM Side Y
Due to osmosis: - A solution containing \(10 \, \text{g}\) of an electrolyte \(AB_2\) in \(100 \, \text{g}\) of water boils at \(100.52^\circ \text{C}\). The degree of ionization of the electrolyte (\(\alpha\)) is _____ \(\times 10^{-1}\). (nearest integer)
\([ \text{Given: Molar mass of } AB_2 = 200 \, \text{g mol}^{-1}, \, K_b \, (\text{molal boiling point elevation const. of water}) = 0.52 \, \text{K kg mol}^{-1}, \text{ boiling point of water} = 100^\circ \text{C}; \, AB_2 \text{ ionises as } AB_2 \rightarrow A^{2+} + 2B^- ]\)
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,