What is the new boiling point of a solution made with 35.7 g MgCl 2 and 150 mL w

Question

What is the new boiling point of a solution made with 35.7 g MgCl2 and 150 mL water? (density of water is 1 g/mL)

a. 1.3°C

b. 101.3°C

c. 3.9°C

d. 103.9°C

What is the vapor pressure of a solution of 2.50 mol of glycerine (non-volatile liquid) in 5.50 mol water at 20°C (vapor pressure of water is .023 atm at 20°C)?

a. .023 atm

b. .016 atm

c. .012 atm

d. none of these

Which of the following statements is true?

a. molarity changes with temperature change, molality does not

b. molality changes with temperature change, molarity does not

c. both molarity and molality change with temperature change

d. neither molarity nor molality change with temperature change

Explanation / Answer

1)

Lets calculate molality first

Molar mass of MgCl2 = 1*MM(Mg) + 2*MM(Cl)

= 1*24.31 + 2*35.45

= 95.21 g/mol

mass of MgCl2 = 35.7 g

we have below equation to be used:

number of mol of MgCl2,

n = mass of MgCl2/molar mass of MgCl2

=(35.7 g)/(95.21 g/mol)

= 0.375 mol

mass of solvent = 150 g (since density is 1g/mL and volume is 150 mL)

= 0.15 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(0.375 mol)/(0.15 Kg)

= 2.4997 molal

i for MgCl2 = 3 (it breaks 1 Mg2+ and 2 Cl-)

lets now calculate deltaTb

deltaTb = i*Kb*m

= 3.0*0.512*2.4997

= 3.8396 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 3.8396

= 103.9 oC

Answer: d

Only 1 question at a time please

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.