When 20g of glucovanillin(an organic compound found in the green fruit of vanilla) are dissolved in 200g..?
a.300 g/mol
b.617 g/mol
c.33.3g/mol
d.265 g/mol
Answer:
Glucovanillin, an organic compound found in the green fruit of vanilla, I assume, is the sugar content, so I would say A?
You need to look three more things up to be able to answer this question:
1) The normal boiling point of (pure) phenol
2) The ebullioscopic constant for phenol. This is a number corresponding to the amount by which the boiling point of phenol is lowered, per mole of dissolved substance.
3) Density of phenol, to calculate the volume of 200 g phenolic solution. Unless you're told/know otherwise, just assume that the solution does not have a volume very different from pure phenol.
Say, BP = normal boiling point and K= ebullioscopic constant, then
(183 - BP) = K* concentration(in moles)
From this, you can compute the molar concentration of glucovanillin. Call it "M".
Convert 200g of phenol to V liters phenol (divide weight by density, keep an eye on the units). Then compute a w/v (weight per volume) concentation, C:
C = 20 g/ V liters.
Now if you divide C by M, you'll have a number with the units "grams per mole", which is what you're after.
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