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QJEEM23Ch744e: JEE Main 2023 Chemistry - Class 12 JEEM23Ch744e Solution: of is required to coagulate of solution in 2 hours. The…

20 mL20 \mathrm{~mL} of 0.5M NaCl0.5 \mathrm{M} ~\mathrm{NaCl} is required to coagulate 200 mL200 \mathrm{~mL} of As2S3\mathrm{As}_{2} \mathrm{S}_{3} solution in 2 hours. The coagulating value of NaCl\mathrm{NaCl} is ____________ .

Step-by-Step Verified 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 mL20\ \mathrm{mL} of 0.5 M0.5\ \mathrm{M} NaCl\mathrm{NaCl} solution is required to coagulate 200 mL200\ \mathrm{mL} of As2S3\mathrm{As}_2\mathrm{S}_3 solution in 2 hours.

To find the coagulating value of NaCl\mathrm{NaCl} , we need to determine the amount of NaCl\mathrm{NaCl} required to coagulate 1 liter (1000 mL) of the As2S3\mathrm{As}_2\mathrm{S}_3 solution in 2 hours. This can be calculated using the following formula:

Coagulating value=amount of electrolyte required to coagulate 1 L of sol in 2 hours1 L

To find the amount of NaCl\mathrm{NaCl} required to coagulate 200 mL of the As2S3\mathrm{As}_2\mathrm{S}_3 solution in 2 hours, we can use the formula for the number of moles of NaCl\mathrm{NaCl} used:

nNaCl=CNaCl×VNaCl

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

nNaCl=0.5 M×0.02 L=0.01 mol

To find the amount of NaCl\mathrm{NaCl} required to coagulate 1 L of the As2S3\mathrm{As}_2\mathrm{S}_3 solution, we need to scale up the amount of NaCl\mathrm{NaCl} used by a factor of 5 (since 5 times the given volume of NaCl\mathrm{NaCl} solution is required to make 1 L of the As2S3\mathrm{As}_2\mathrm{S}_3 solution). This gives:

nNaCl=0.01 mol×5=0.05 mol

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

Coagulating value=0.05 mol1 L×1000=50 mM

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

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