Question

List the following aqueous solutions in order of increase in boiling point:

a. 0.1 m NaCl,
b. 0.1 m Cu(NO3)2,
c. 0.1 m C6H12O6,
d. 0.1 m Fe(NO3)3

Answer 1

0.1 m C6H12O6 < 0.1 m NaCl < 0.1 m Cu(NO3)2 < 0.1 m Fe(NO3)3

Step-by-step explanation:

Step 1: Data given

ΔTb = Kb*m*i

⇒with Kb = 0.512°C/m

⇒with m = molality

⇒with i = Van't hoff factor = the number of particles into which the solute dissociates

a. 0.1 m NaCl

⇒ molality = 0.1 molal

⇒ i(NaCl) = 2 ( dissociates in Na+ and Cl-)

0.1 * 2 = 0.2

b. 0.1 m Cu(NO3)2

⇒ molality = 0.1 molal

⇒ i(Cu(NO3)2) = 3 ( dissociates in Cu^2+ and 2NO3-)

0.1 * 3 = 0.3

c. 0.1 m C6H12O6

⇒ molality = 0.1 molal

⇒ i(C6H12O6) = 1 ( doesn't dissociate in water)

0.1 * 1 = 0.1

d. 0.1 m Fe(NO3)3

⇒ molality = 0.1 molal

⇒ i(Fe(NO3)3) = 4 ( dissociates in Fe^3+ and 3NO3-)

0.1 * 4 = 0.4

0.1 m C6H12O6 < 0.1 m NaCl < 0.1 m Cu(NO3)2 < 0.1 m Fe(NO3)3