There are two ways to solve this problem.
Method 1
Use the Ideal Gas Law to calculate the number of moles.
#PV = nRT#
#n = (PV)/(RT)#
#P = 714color(red)(cancel(color(black)("Torr"))) × "1 atm"/(760color(red)(cancel(color(black)("Torr")))) = "0.9395 atm"#
#V = "114 mL" = "0.114 L"#
#T = "(33+273.15) K" = "306.2 K"#
#n = (0.9395 color(red)(cancel(color(black)("atm"))) × 0.114color(red)(cancel(color(black)("L"))))/("0.082 06"color(red)(cancel(color(black)("L·atm·K"^"-1""mol"^"-1"))) × 306.2color(red)(cancel(color(black)("K")))) = "0.004 263 mol"#
#"Molar mass" = "grams"/"moles" = "0.170 g"/"0.004 263 mol" = "39.9 g/mol"#
Method 2
Rearrange the Ideal Gas Law to do the calculation all in one step.
#PV = nRT = g/MRT#, where #g# is the mass and #M# is the molar mass.
#M = (gRT)/(PV) = (0.170"g" × "0.082 06"color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 306.2color(red)(cancel(color(black)("K"))))/(0.9395color(red)(cancel(color(black)("atm"))) × 0.114color(red)(cancel(color(black)("L")))) = "39.9 g/mol"#
