A 2.900×102 M solution of NaCl in water is at 20.0C. The sample was created by d

Question

A 2.900×102 M solution of NaCl in water is at 20.0C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.2 mL . The density of water at 20.0C is 0.9982 g/mL.

Part A Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units.

Part B Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures.

Part C Calculate the concentration of the salt solution in percent by mass. Express your answer to four significant figures and include the appropriate units.

Part D Calculate the concentration of the salt solution in parts per million. Express your answer as an integer to four significant figures and include the appropriate units.

Explanation / Answer

part A

m = M*1000/(D*1000 - M*MWT)

m = molality = ?

M = molarity = 0.29 M

D = density = 0.9982 g/ml

Mwt of solute (NaCl) = 58.5 g/mol

m = (0.29*1000)/(0.9982*1000 - 0.29*58.5) = 0.2955 m

part B

mole fraction of salt = no of mol of salt/total no of mol

no of mol of salt = M*V = 0.29*1 = 0.29 mol

no of mol of water = wt/mwt = (999.2*0.9982)/18 = 55.41 mol

mole fraction of salt = 0.29/(55.41+0.29) = 0.005206

part C

percent by mass = wt of solute / wt of solution*100

wt of solute (NaCl) = 0.29*58.5 = 16.965 grams

wt of solution = 16.965+997.4 = 1014.4 grams

% bymass = 16.965/1014.4*100 = 1.672%

part D

PPM = WT of solute /wt of solution*10^6

= (16.965/1014.4)*10^6

= 1.672*10^4 ppm

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