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).

Surface Chemistry

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