pure water bols at 1000 C, and sin the adtion of solute noreases boiling point t
ID: 104705 • Letter: P
Question
pure water bols at 1000 C, and sin the adtion of solute noreases boiling point the boling point f an aquecus solution, T ez ng port of an aqueous solution, Tr, will be Since pure water sat 0.0 and s nehe addition of solute de re ses eezing por e o om the T,-(0.00-dri) °C Part A are the stants What is the bolling point of a solution made using 845 g of sucrose, CaH2O11, in 0.225 kg of water, H20? Express your answer in degrees Celsius using five significant figures Hints Submit My Answers Give Up Part B What is the freezing point of a solution that contains 28.0 g af urea, CO(NH2)a in 295 ml, water. H,07 Assume a densilty of water of 1,00 g/ml Express your answer numerically in degrees Celsius Hints Submit My Answers Give Up Provide FeedbackExplanation / Answer
A)
Lets calculate molality first
Molar mass of C12H22O11,
MM = 12*MM(C) + 22*MM(H) + 11*MM(O)
= 12*12.01 + 22*1.008 + 11*16.0
= 342.296 g/mol
mass(C12H22O11)= 845 g
use:
number of mol of C12H22O11,
n = mass of C12H22O11/molar mass of C12H22O11
=(845.0 g)/(342.296 g/mol)
= 2.469 mol
m(solvent)= 0.225 Kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(2.469 mol)/(0.225 Kg)
= 10.97 molal
lets now calculate Tb
Tb = Kb*m
= 0.512*10.9717
= 5.6175 oC
This is increase in boiling point
boiling point of pure liquid = 100.0 oC
So, new boiling point = 100 + 5.6175
= 105.6175 oC
Answer: 105.62 oC
B)
Lets calculate molality first
Molar mass of CO(NH2)2,
MM = 1*MM(C) + 1*MM(O) + 2*MM(N) + 4*MM(H)
= 1*12.01 + 1*16.0 + 2*14.01 + 4*1.008
= 60.062 g/mol
mass(CO(NH2)2)= 28.0 g
use:
number of mol of CO(NH2)2,
n = mass of CO(NH2)2/molar mass of CO(NH2)2
=(28.0 g)/(60.062 g/mol)
= 0.4662 mol
m(solvent)= 295 g
= 0.295 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.4662 mol)/(0.295 Kg)
= 1.58 molal
lets now calculate Tf
Tf = Kf*m
= 1.86*1.5803
= 2.9393 oC
This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 2.9393
= -2.9393 oC
Answer: -2.94 oC
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.