$20 \mathrm{~mL}$ of $0.5 \mathrm{M} ~\mathrm{NaCl}$ is required to coagulate $200 \mathrm{~mL}$ of $\mathrm{As}_{2} \mathrm{S}_{3}$ solution in 2 hours. The coagulating value of $\mathrm{NaCl}$ is ____________ .
Answer (integer)
50
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.
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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:
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$$\text{Coagulating value} = \frac{\text{amount of electrolyte required to coagulate 1 L of sol in 2 hours}}{1\ \mathrm{L}}$$
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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:
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$n_{\mathrm{NaCl}} = C_{\mathrm{NaCl}} \times V_{\mathrm{NaCl}}$
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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:
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$n_{\mathrm{NaCl}} = 0.5\ \mathrm{M} \times 0.02\ \mathrm{L} = 0.01\ \mathrm{mol}$
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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:
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$n_{\mathrm{NaCl}} = 0.01\ \mathrm{mol} \times 5 = 0.05\ \mathrm{mol}$
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The coagulating value of $\mathrm{NaCl}$ can now be calculated as:
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$$\text{Coagulating value} = \frac{0.05\ \mathrm{mol}}{1\ \mathrm{L}} \times 1000 = 50\ \mathrm{mM}$$
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Therefore, the coagulating value of $\mathrm{NaCl}$ is $\boxed{50}$ (in millimoles per liter).
About this question
Subject: Chemistry · Chapter: Surface Chemistry · Topic: Adsorption
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