A 0.320 m ( molality ) aqueous solution of NaCl is prepared at 20.0C. Assume tha
ID: 918819 • Letter: A
Question
A 0.320 m (molality) aqueous solution of NaCl is prepared at 20.0C. Assume that the density of the solution at 20.0C is 1.082 g/mL.
A.) Calculate the molarity of the salt solution.
Express your answer to four significant figures and include the appropriate units.
B.) Calculate the mole fraction of salt in this solution.
Express the mole fraction to four significant figures.
C.) Calculate the concentration of the salt solution in percent by mass. (percent by mass NaCl)
Express the mass percent to three significant figures.
Please include work if you can! Thanks!
Explanation / Answer
m = mol NaCl / kg water
a)
M = mol/L
V = 1.082*1000 = 1082 L
mol = 0.32
M = 0.32/1082 = 0.000295 M o fNaCl
b)
%mol frac = mol NaCl / Total mol
mol NaCl = 0.32 mol
mol H2O = mass/MW = 1000/18 = 55.555
mol frac Nacl = 0.032/(55.555+0.032) = 0.0005756
C)
% mass = mass NaCl / total mass * 100
mass of NaCl = mol*MW = 0.32*58 = 18.56
total mass = 1000 + 18.56 = 1018.56
Then
% mass = 18.56 / (1018.56)*100 = 1.822%
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