Question

A 6.00  g sugar cube (sucrose: C12H22O11) is dissolved in a 350 mL teapot containing 80∘ C water (density of water at 80∘ C= 0.975 g/mL). What is the molality of the sugar solution?

Express the molality with the appropriate units.

Calcium chloride, CaCl2, is commonly used as an electrolyte in sports drinks and other beverages, including bottled water. A solution is made by adding 8.50  g of CaCl2 to 60.0 mL of water at 25∘C. The density of the solvent at that temperature is 0.997 g/mL. Calculate the mole percent of CaCl2 in the solution.

Express your answer as a percent to three significant figures.

Answer 1

1)Moles of sugar in solution=6.00g/molar mass of sugar=6.00g/342g/mol=0.0175 mol

[C12H22O11=12*12+22*1+16*11=144+22+176=342 g/mol]

Now density of water=0.975g/ml

So mass of water in 350ml=0.975 g/ml*350ml=341.25 g=341.25 /1000 =0.34125 kg

So molality of sugar solution=moles of solute/kg of solvent

                                                   =0.0175 mol sugar/0.34125kg =0.0513 mol/kg

2) moles of Cacl2=8.50g/molar mass of Cacl2=8.50g/111g/mol=0.0766 mol

[CaCl2=40+35.5*2=111g/mol]

Mass of water in solution =density of water*volume=0.997g/ml*60.0ml=59.82 g

Moles of water in solution=59.82g water/molar mass of water=59.82g/18g/mol=3.323 mol

Mole percent of Cacl2=(moles of Cacl2 in solution /total moles in solution)*100=0.0766 mol/0.0766 mol+3.323 mol)*100=(0.0766/3.3996)*100=2.253%