What is the boiling point of a solution of 1.20g of glycerol, C3H8O3, in 20.0 g of water? What is the freezing point?

TB solution = 100.3315 °C

T freezing solution = -1.209 °C

Given

1.2 g glycerol

20 g water

Required

the boiling point

the freezing point

Solution

ΔTB = KB . m

ΔTF = KF . m

mass water = 20 g = 0.02 kg

Kb water = 0.51 °C/m

KF water = 1.86 °C/m

mol glycerol :

= 1.2 g : 92,09382 g/mol

= 0.013

molal  : the number of moles of solute in 1 kg of solvent  

molal of solution :

= 0.013 : 0.02

= 0.65 m

ΔTB = KB . m

ΔTB = TB solution - TB solvent  

TB solution - 100 = 0.51 x 0.65

TB solution - 100 = 0.3315

TB solution = 100.3315 °C

ΔTF = KF . m

ΔT = KF . m

ΔTF = 1.86 x 0.65

ΔTF = 1.209

T_f  = -1.209 °C

screening

explanation:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol