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.saplinglearning.com/ibiscms/mod/ibis/view.php?id 3594600 West Virginia Northern Community College- CHEM 108- Summer17- BROWN Activities and Due Dates Final Exam 6/30/2017 1 1:55 PM Gradebo Periodic T ble Question 9 of 55 Mop Sapling Learning Your lab partner combined chloroform (CHCla) and acetone (C,HO) to create a solution where the mole fraction of chloroform, Xctirom is 0.187. The densities of chloroform and acetone are 1.48 g/mL and 0.791 g/mL, respectively Calculate the molarity of the solution. Number Calculate the molality of the solution. Number Next Savo And Exit O Previous about us careers privac Learning, Inc DOL

Explanation / Answer

x1 = 0.187

x2 = 1-x1 = 1-0.187 = 0.813

a)

M = mol of chloroform / Volume of solution

Frist, assume a basis, let it be 1 mol of solution

so,

mol of chloroform = 0.187

mol of acetone = 0.813

now,

mass of chloroform = mol*MW = 0.187 *119.38 =22.324 g

mass of acteone = mol*MW = 0.813 *60 = 48.78 g

change to volume, and assume volumes are additive

V of chlorofom = mass/Density = 22.324 / 1.48 = 15.083783 mL

V of acetone = 48.78 /0.791 = 61.66 mL

Total V = 61.66+15.083783 = 76.743 mL = 76.743 *10^-3 L

now,

M = mol of solute / Total V

mol of solute = 0.187 / ( 76.743 *10^-3)= 2.4367M

b)

molality = mol of solute / kg of solvent

mol of solute = 0.187 mol

kg of solvent = 48.78 g = 48.78 *10^- 3 kg

molality = 0.187 /(48.78 *10^-3) = 3.83353 molal

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