Question

Calculate the mass percent, molality and mole fraction of citric acid (H_3C_6H_5O_7) in a 1.37 M aqueous solution which has a density of 1.10 g/cm^3. (Assume a 1.00 L solution.) Citric acid is triprotic.

Answer 1

To find the %mass of citric acid we need to calculate the moles

$1.37 \frac{mol C.A.}{1L}\frac{192 g C.A.}{1 mol C.A.}*1L=16.44 g C.A.$

$1.10\frac{g, sln}{cm3}\frac{100^{3}cm3}{1 m3}\frac{1 m3}{1000L}*1L=1100 g, sln$

$perc, mass=\frac{mass, of C.A.}{mass, of sln}*100=\frac{16.44}{1100}*100=1.54$

Mass percent of C.A.= 1.54%

To find the molality we use the following equation:

$molality=\frac{mol solute}{kg dissolvent}=\frac{1.37 mol, c.a.}{1.1 kg, sln}=1.245$

To find the moles of the solution we need the %m, H2O and %m of citric acid

Solving for x we get:

x=10832.46 g H2O

Then we obtain the amount of moles of the disolution.

$16.54g C.A.\frac{1 mol}{192g,CA.}+1083.46g,H2O\frac{1 mol H2O}{18 g,H2O}=60.28 mol, sln$

Then we divide the mol of C.A. between the mol of the soluton to obtain the %mol

$\frac{1.37mol C.A.}{60.27mol,sln}*100=2.27$

=2.27% mol