Question #fbb60
1 Answer
Here's what I got.
Explanation:
Judging by the information you provided, I would assume that you're actually looking for the molality of the resulting solution, not for its molarity.
I say this because in order to calculate molarity, you would need to know the volume of the resulting solution. In this case, you would need to calculate the percent concentration by mass of the solution, then look up its corresponding density.
Molality, on the other hand, requires moles of solute, which in your case is vanillin, divided by kilograms of solvent, which in your case is diphenyl ether.
#color(blue)("molality" = "moles of solute"/"kilograms of solvent")#
To get the number of moles of solute, use vanillin's molar mass, which tells you what the mass of one mole of the compound is.
Don't forget to convert the mass of the sample from milligrams to grams!
#39.1color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(1000color(red)(cancel(color(black)("mg")))) * ("1 mole C"_8"H"_8"O"_3)/(152.15color(red)(cancel(color(black)("g")))) = "0.000257 moles C"_8"H"_8"O"_3#
Convert the mass of the solvent from milligrams to kilograms
#168.5color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(1000color(red)(cancel(color(black)("mg")))) * "1 kg"/(1000color(red)(cancel(color(black)("g")))) = 168.5 * 10^(-6)"kg"#
The molality of the solution will thus be
#color(blue)(b = n_"solute"/m_"solvent")#
#b = "0.000257 moles"/(168.5 * 10^(-6)"kg") = color(green)("1.53 molal")#
