JEE Main 2023 — Surface Chemistry Question with Solution

The coagulating value of an electrolyte is defined as the minimum amount of the electrolyte required to coagulate a unit volume of the colloidal solution in a given time. In this problem, we are given that $20\mathrm{mL}$ of $0.5\mathrm{M}$ $\mathrm{NaCl}$ solution is required to coagulate $200\mathrm{mL}$ of ${\mathrm{As}}_2{\mathrm{S}}_3$ solution in 2 hours.

To find the coagulating value of $\mathrm{NaCl}$, we need to determine the amount of $\mathrm{NaCl}$ required to coagulate 1 liter (1000 mL) of the ${\mathrm{As}}_2{\mathrm{S}}_3$ solution in 2 hours. This can be calculated using the following formula:

$$\text{Coagulating value}=\frac{\text{amount of electrolyte required to coagulate 1 L of sol in 2 hours}}{1\mathrm{L}}$$

To find the amount of $\mathrm{NaCl}$ required to coagulate 200 mL of the ${\mathrm{As}}_2{\mathrm{S}}_3$ solution in 2 hours, we can use the formula for the number of moles of $\mathrm{NaCl}$ used:

$$n_{\mathrm{NaCl}}=C_{\mathrm{NaCl}}\times V_{\mathrm{NaCl}}$$

where $C_{\mathrm{NaCl}}$ is the concentration of $\mathrm{NaCl}$ in moles per liter, and $V_{\mathrm{NaCl}}$ is the volume of $\mathrm{NaCl}$ solution used in liters. Substituting the given values, we get:

$$n_{\mathrm{NaCl}}=0.5\mathrm{M}\times 0.02\mathrm{L}=0.01\mathrm{mol}$$

To find the amount of $\mathrm{NaCl}$ required to coagulate 1 L of the ${\mathrm{As}}_2{\mathrm{S}}_3$ solution, we need to scale up the amount of $\mathrm{NaCl}$ used by a factor of 5 (since 5 times the given volume of $\mathrm{NaCl}$ solution is required to make 1 L of the ${\mathrm{As}}_2{\mathrm{S}}_3$ solution). This gives:

$$n_{\mathrm{NaCl}}=0.01\mathrm{mol}\times 5=0.05\mathrm{mol}$$

The coagulating value of $\mathrm{NaCl}$ can now be calculated as:

$$\text{Coagulating value}=\frac{0.05\mathrm{mol}}{1\mathrm{L}}\times 1000=50\mathrm{mM}$$

Therefore, the coagulating value of $\mathrm{NaCl}$ is $\boxed{50}$ (in millimoles per liter).