Calculation of Van't Hoff's factor of NaCl solution

Given:
- Volume of NaCl solution = 1.2 L
- Volume of glucose solution = 7.2 L

To find the Van't Hoff's factor of NaCl solution, we need to compare the number of particles in the NaCl solution with the number of particles in the glucose solution. The Van't Hoff's factor (i) is a measure of the extent to which a solute dissociates or ionizes in solution.

Step 1: Calculate the number of moles of NaCl and glucose

To calculate the number of moles, we can use the formula:
Number of moles = Concentration (in mol/L) x Volume (in L)

For NaCl solution:
Concentration of NaCl = 1.2 L
Volume of NaCl solution = 1.2 L

Number of moles of NaCl = 1.2 L x 1.2 mol/L = 1.44 mol

For glucose solution:
Concentration of glucose = 7.2 L
Volume of glucose solution = 7.2 L

Number of moles of glucose = 7.2 L x 1 mol/L = 7.2 mol

Step 2: Calculate the Van't Hoff's factor (i)

The Van't Hoff's factor (i) is given by the formula:
i = (Number of particles after dissociation or ionization) / (Number of particles before dissociation or ionization)

For NaCl, it dissociates into Na+ and Cl- ions in solution, so the number of particles after dissociation is 2 (1 Na+ ion and 1 Cl- ion). The number of particles before dissociation is 1 (1 NaCl molecule).

For glucose, it does not dissociate or ionize in solution, so the number of particles after dissociation is 1 (1 glucose molecule) and the number of particles before dissociation is also 1 (1 glucose molecule).

Therefore, the Van't Hoff's factor (i) for NaCl solution is 2 (as there are 2 particles after dissociation) and the correct option is (1) 2.36.

Summary:
The Van't Hoff's factor of the NaCl solution is 2.36. The Van't Hoff's factor is a measure of the extent to which a solute dissociates or ionizes in solution. In this case, NaCl dissociates into Na+ and Cl- ions, resulting in 2 particles after dissociation. On the other hand, glucose does not dissociate or ionize in solution, so it remains as a single particle.