20 mL of 0.5 M NaCl is required to coagulate 200 mL of As_2 S_3 solution in 2 hours. The coagulating value of NaCl is .

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Correct answer: 50. (50) 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 mL of 0.5 M NaCl solution is required to coagulate 200 mL of As_2 S_3 solution in 2 hours. To find the coagulating value of NaCl, we need to determine the amount of NaCl required to coagulate 1 liter (1000 mL) of the As_2 S_3 solution in 2 hours. This can be calculated using the following formula: Coagulating value= amount of electrolyte required to coagulate1 L of sol in2 hours1 L To find the amount of NaCl required to coagulate 200 mL of the As_2 S_3 solution in 2 hours, we can use the formula for the number of moles of NaCl used: n_ NaCl=C_ NaCl V_ NaClwhere C_ NaCl is the concentration of NaCl in moles per liter, and V_ NaCl is the volume of NaCl solution used in liters. Substituting the given values, we get: n_ NaCl=0.5M 0.02 L=0.01 mol To find the amount of NaCl required to coagulate 1 L of the As_2 S_3 solution, we need to scale up the amount of NaCl used by a factor of 5 (since 5 times the given volume of NaCl solution is required to make 1 L of the As_2 S_3 solution). This gives: n_ NaCl=0.01 mol 5=0.05 mol The coagulating value of NaCl can now be calculated as: Coagulating value = 0.05 mol1 L 1000=50 mM Therefore, the coagulating value of NaCl is 50 (in millimoles per liter).

$20 mL of 0.5 M NaCl is required to coagulate 200 mL of As_{2} S_{3} solution in 2 hours. The coagulating value of NaCl is .$

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