What is the concentration of a solution containing 1.11 g sugar (sucrose, c12h22o11, mw = 342.3 g/mol, d = 1.587 g/cm3) in 432 ml glycerine (c3h8o3, mw = 92.09 g/mol, d = 1.261 g/cm3) in molarity and molality.

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph