A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Question

lovelyharmonics · Answer

@Kainui

lovelyharmonics · Answer

i was reading through the mathmatics chat box and at first i thought they were speaking a different language.... it turns out that they're all just trying to talk ghetto.....

jfraser · Answer

do you know the equation for freezing-point depression?

lovelyharmonics · Answer

yus c: the freezing point comes from solvent x mol x ions

jfraser · Answer

not exactly. Try this:

$$\Delta T_f = \frac{moles \space solute}{kg \space solvent}*K_f$$

lovelyharmonics · Answer

.-. at least i got all the components right :D

jfraser · Answer

but if they're in the wrong place you'll never get it.

jfraser · Answer

you have all the pieces that the equation needs, almost

jfraser · Answer

you have grams of solute, you need to find the moles

jfraser · Answer

you have grams of solvent, you need kilograms

jfraser · Answer

and you have the Kf factor

lovelyharmonics · Answer

so do i just multiply 21.5 by 6.022x10^23?

lovelyharmonics · Answer

@ikram002p

abhisar · Answer

Hello @lovelyharmonics 

$\sf \Delta T_f=K_f\times molality(\frac{\large moles~of~solute}{\large mass~of~solvent~in~kg})$

abhisar · Answer

molar mass of glucose = 180g
=> moles of glucose in 21.5 grams=21.5/180=0.119
=> Molality = 0.119/0.255=0.46

Now,
$\sf \Delta T_f=-1.86\times 0.46=0.87°C$

abhisar · Answer

Got it ?