98.7 mL CH3CN x 0.786 g/mL = about 77.6g
mols CH3CN = 77.6/41 = about 1.9
22.5 mL CH3OH x 0.791 = estimated about 18 g
mols CH3OH = about 18/32 = about 0.56
a. XCH3OH = mols CH3OH/total mols
b. There is no molality of the solution. Since you're asking for molarity CH3OH in part c I'll assume you want molality of CH3OH.
molality = m = mols CH3OH/kg CH3CN
c. M CH3OH = mols CH3OH/L of solution. volume of solution will be 22.5 mL + 98.7 mL = 121.2 mL assuming they are additive.
Note that I have estimated al along. You should redo each calculation and use your values.
Post your work if you get stuck.
The density of acetonitrile (CH3CN) is 0.786g /mL and the density of methanol (CH3OH) is 0.791g/mL. A solution is made by dissolving 22.5 mL of CH3OH in 98.7 mL of CH3CN . (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of CH3OH in the solution?
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