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# A mixture of NaCl and sucrose (C12H22O11)...?

A mixture of NaCl and sucrose (C12H22O11) of combined mass 10.2 g is dissolved in enough water to make up a 250 mL solution. The osmotic pressure of the solution is 7.09 atm at 23°C. Calculate the mass percent of NaCl in the mixture

### 1 Answer

- Roger the MoleLv 72 years ago
Let z be the mass (in grams) of the NaCl.

Then 10.2 - z is the mass of sucrose.

The NaCl dissociates completely in water, so its van't Hoff factor is 2.

(z g NaCl) / (58.4430 g NaCl/mol) x (2 mol ions / 1 mol NaCl) = 0.0342214 z mol ions from NaCl

((10.2 - z) g C12H22O11) / (342.2965 g C12H22O11/mol) =

(0.0297987 - 0.00292144 z) mol C12H22O11, which does not dissociate

n = PV / RT = (7.09 atm) x (0.250 L) / ((0.08205746 L atm/K mol) x (23 + 273) K) = 0.072975 mol total solutes

Set the sum of the two values for moles of solutes equal to the total moles just found:

0.0342214 z + (0.0297987 - 0.00292144 z) = 0.072975

Solve for z algebraically:

z = 1.3794 g NaCl

(1.3794 g NaCl) / (10.2 g total) = 0.135 = 13.5% NaCl by mass