1. The vapor pressure of pure water at 85oC is 434 torr. What is the

vapor pressure at 85oC of a solution prepared from 100g of water and 150 g of diglyme, C6H1403?

360.22 Torr

Explanation:

In this case, we need to use the Raoult Law expression for vapour pressure which is the following:

Ps = Xw * P°w

Ps: Pressure of solution

Xw: molar fraction of water

P°w: vapour pressure of water.

Of course this expression is only usable in cases like this, when we have solutions with non volatile solutes like this one.

Now, we have the pressure of water but not the molar fraction. This can be calculated using the masses and then, calculating the moles.

the moles of each reagent here, using the molar masses, which are 18 g/mol for water and 134 g/mol for dyglime:

nw = 100 / 18 = 5.56 moles

nd = 150 / 134 = 1.12 moles

Now the molar fraction can be done using the following expression:

X = na / na + nb

We have the moles of each compound, so, the molar fraction for water it would be:

Xw = 5.56 / (5.56+1.12) = 0.83

Finally, we can calculate the vapour pressure of solution Ps:

Ps = 0.83 * 434

Ps = 360.22 Torr

c. 75.7

i think that is the right