Question

1. Estimate the boiling point of the solution in degrees Celsius consisting of 19.5g of glucose (M=180.16 g/mol) dissolved in 120g of water; having Kf = 1.86 K kg mol-1 and Kb = 0.51 K kg mol-1 for water.

2. What is the concentration of glucose in an aqueous solution which exhibits an osmotic pressure of 16 mm of water at 25^oC. State your answer in mM.

Answer 1

1. Estimate the boiling point of the solution in degrees Celsius consisting of 19.5g of glucose (M=180.16 g/mol) dissolved in 120g of water; having Kf = 1.86 K kg mol-1 and Kb = 0.51 K kg mol-1 for water.

Solution :-

Lets first calculate the molality of the solution

Moles of glucose = mass / molar mass

                            = 19.5 g / 180.16 g per mol

                            = 0.1082 mol

Molality = moles of solute / kg solvent

             = 0.1082 mol / 0.120 kg

            = 0.902 m

Now using the Kb and molality we can find the change in the boiling point

Tb= Kb* m

             = 0.51 ^oC kg mol-1 * 0.902 mol per kg

             = 0.46 ^oC

Boiling point of solution = boiling point of pure water + Tb

                                      = 100 C + 0.46 C

                                      = 100.46 ^oC

Hence the boiling point of the solution is 100.46 ^oC or we can round it up to 100.5 ^oC

2. What is the concentration of glucose in an aqueous solution which exhibits an osmotic pressure of 16 mm of water at 25^oC. State your answer in mM.

Solution :- formula to calculate the osmotic pressure is as follows

= MRT

where, = osmotic pressure

M= molarity

R= gas constant (0.08206 L atm per mol K)

T=Kelvin temperature (25 + 273 = 298 K)

Lets convert 16 mm water to atm

16 mm water * 1 mmHg / 13.56 mm water = 1.18 mmHg

1.18 mm Hg *1atm / 760 mmHg = 0.001553 atm

Now lets use it to calculate the concentration of the glucose

Pi = MRT

0.001553 atm = M * 0.08206 L atm per mol K * 298 K

0.001553 atm /( 0.08206 L atm per mol K * 298 K) = M

6.35*10^-5 M= M

lets convert the M to mM

6.35*10^-5 M * 1000 mM /1 M = 6.35*10^-2 mM

Therefore the concentration of the glucose is 6.35*10^-2 mM