A solution is made by dissolving 19.5 grams of glucose (C6H12O6) in .245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting freezing point of the solution? Show all of the work needed to solve this problem.

Question

superhelp101 · Answer

@Nurali

ganeshie8 · Answer

I think we need to use this equation
$$\large \Delta T_f = iK_f m_{solute}$$

superhelp101 · Answer

yes we do

ganeshie8 · Answer

$i = 1$ for glucose
you're given $K_f = -1.86^{\circ}C/m $

ganeshie8 · Answer

know how to calculate the molality $\large  m_{solute }$ ?

superhelp101 · Answer

molarity = moles of solute/ volume of solution in litres
@chmvijay

chmvijay · Answer

Molalaity = moles /litre of solution
Molality= moles /Kg of solution

superhelp101 · Answer

don't i have to convert something to moles?

chmvijay · Answer

19.5 grams of glucose

chmvijay · Answer

moles = gram/molar mass :P

superhelp101 · Answer

do i just plug 19.5 in the equation?

chmvijay · Answer

which equation lol 
moles of glucose=19.5 / molar mass of glucose

superhelp101 · Answer

ohhh ok 
19.5/ 180.1559

chmvijay · Answer

yaaa its mole now u calculate MOLALITY

superhelp101 · Answer

.10823959

chmvijay · Answer

lol what is this :P

superhelp101 · Answer

19.5/ 180.1559= .10823959 :)

chmvijay · Answer

itsa mole of glucose u wanted to find molality right ? 
molality = moles of glucose  /kg of solution

superhelp101 · Answer

wait i am so confused? :(

chmvijay · Answer

first u have to find moles of glucose then u have to find out the molality of glucose solution :P

chmvijay · Answer

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