Question

18 g of glucose, CH206 is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? K, for water is 0.52 K kg mor!​

Answer 1

Answer:

Given that

Weight of the glucose is 18g

Mass of the glucose is 180g

K

b

for water is 0.52 Kkg/mol

As we know,

At 1 atm, the boiling point of water is 100 degree celcius.

ΔT

b

=

(wt of water×GMW of glucose)

(K

b

×wt of glucose)

ΔT

b

=

180×1

0.52×18

ΔT

b

=

180

9.36

ΔT

b

=0.052

that implies

T

S

=0.052+373.15 = 373.202

T

S

= 373.202 K

Therefore water boiling point is 373.202 K..