Arrange the following in the increasing order of their van’t Hoff factors:

(A) Very dilute \(MgCl_2\) solution

(B) Very dilute \(AlCl_3\) solution

(C) Very dilute \(NaCl\) solution

(D) Very dilute \(Al_2O_3\) solution

(E) Very dilute Urea solution

Choose the correct answers from the options given below:

(E) < (C) < (A) < (B) < (D)

The correct answer is option 1. (E) < (C) < (A) < (B) < (D)

The van't Hoff factor (\(i\)) is a measure of the number of ions formed when a solute dissolves in a solvent. For ionic compounds, it is equal to the number of ions produced when the compound dissociates.
Let's consider each of the given substances and determine their van't Hoff factors:

(A) Very dilute \(MgCl_2\) solution:

\(MgCl_2\) dissociates into \(Mg^{2+}\) and \(2Cl^-\), so

\(i = 1 + 2 = 3\).

(B) Very dilute \(AlCl_3\) solution:

\(AlCl_3\) dissociates into \(Al^{3+}\) and \(3Cl^-\), so

\(i = 1 + 3 = 4\).

(C) Very dilute \(NaCl\) solution:

\(NaCl\) dissociates into \(Na^+\) and \(Cl^-\), so \(i = 1 + 1 = 2\).

(D) Very dilute \(Al_2O_3\) solution:

\(Al_2O_3\) dissociates into \(2Al^{3+}\) and \(3O^{2-}\), so \(i = 2 + 3 = 5\).

(E) Very dilute Urea solution:

Urea (\(CO(NH_2)_2\)) is a molecular compound and does not dissociate into ions when it dissolves. Therefore, \(i = 1\) (for the molecule as a whole).

Now, let's arrange them in increasing order of their van't Hoff factors:

1. Very dilute Urea solution (E) with \(i = 1\).

2. Very dilute \(NaCl\) solution (C) with \(i = 2\).

3. Very dilute \(MgCl_2\) solution (A) with \(i = 3\).

4. Very dilute \(AlCl_3\) solution (B) with \(i = 4\).

5. Very dilute \(Al_2O_3\) solution (D) with \(i = 5\).

So, the correct order is: (1) (E) < (C) < (A) < (B) < (D)