Question

How many grams of glucose, C6H12O6, are required to lower the freezing point of 150g of water by 0.75 degrees celcius? What will be the boiling point of the solution?

Answer 1

@shrutipande9

Answer 2

@thomaster

Answer 3

can you help me with this assignment :D

Answer 4

Kf for water is 1.86 deg C/m

Answer 5

what is the working formula?

Answer 6

\[DeltaTf = Kf * m\]
0.75 = 1.86 * moles of solute/weight of solvent

Answer 7

can u put the necessary values?

Answer 8

u are 1 step ahead of me..:D
great..:) now can u find the value of W of solute

Answer 9

r u there? @~Gelmhar

Answer 10

11g

Answer 11

now the boiling point :D

Answer 12

10.8 to be precise...but yaay u got it...

Answer 13

bec'z i compute 10.89 lol

Answer 14

now elevation in boiling point is \[\Delta Tb = Kb * m\]

Answer 15

0.512 = Kb?

Answer 16

yep

Answer 17

m= 0.41 m

Answer 18

so i will just times, them?

Answer 19

Tbsolvent = 100

Answer 20

so m *o.512

Answer 21

0.21 + 100
so i got 100.21 as the boiling point

Answer 22

yep..:D

Answer 23

we have exam tommorow .. im glad you guide me here LOL

Answer 24

all d best for tomo..:D
u will do good..:)
if u think a user helps u..always appreciate them with a medal..:)

Answer 25

can you help me with another problem :D

Answer 26

sure

Answer 27

I will grand your wish if you help me 1 more time :D

Answer 28

just make a new ques..:)

Answer 29

Is it ok, if i will just type the question here or make another?

Answer 30

rules..:/
out of all i hv to follow

Answer 31

make another is better