Question

Show your work with good use of units, rounding, and significant figures. [Hint: it is good practice to show the value of your answer before you round off to the final answer with the correct significant figures!]

(7 points) A solution was prepared by dissolving 18.00 g of glucose in 150.0 g of water. The molar mass of glucose is 180.15 g/mol. The boiling point elevation constant for water is 0.512 °C•kg/mol.



What is the molality of the solution?
What is the resulting solution's new boiling point?

Answer 1

Answer:Boiling point of a solution is found to be 100.34

o

C. Boiling point of pure

water is 100

o

C.

The elevation in the boiling point ΔT

b

​

=100.34−100=0.34

o

C.

12 gm glucose (molecular weight 180 g/mol) is dissolved in 100 gm water.

The number of moles of glucose =

180g/mol

12g

​

=0.0667mol

Mass of water =100g×

1000g

1kg

​

=0.100kg

Molality of solution m=

0.100kg

0.0667mol

​

=0.667mol/kg

The elevation in the boiling point ΔT

b

​

=K

b

​

×m

0.34

o

C=K

b

​

×0.667mol/kg

K

b

​

=0.51

o

Ckg/mol

The molal elevation constant for water is 0.51

o

Ckg/mol.

Step-by-step explanation: