CHEMISTRY HELP PLEASE!!! A solution is prepared by mixing 1.72 grams of propanol (C3H) in 150g of water.?

a)
Molar mass of propanol (C₃H₇OH) = (12.0×3 + 1.0×8 + 16.0) g/mol = 60.0 g/mol
Molality = (1.72/60.0 mol) / (150/1000 kg water) = 0.191 m

b)
Mass percent of propanol = [1.72/(150+1.72)] × 100% = 1.13%

c)
Molar mass of water (H₂O) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Moles of propanol = 1.72/60 mol
Moles of water = 150/18.0 mol
Mole fraction of propanol = (1.72/60)/[(1.72/60) + (150/18.0)] = 0.00343

d)
Volume of the solution = [(1.72 + 150) g] / (1250 g/L) = 151.72/1250 L
Molarity of propanol = (1.72/60 mol) / (151.72/1250 L) = 0.236 M

There are problems with this question.  Firstly, propanol only describes the empirical formula of the compound in question, although that accuracy is not strictly speaking necessary.  Secondly, that isn't a chemical formula in brackets, and it certainly isn't the formula for any kind of propanol.

A solution is prepared by mixing 1.72 grams of propanol (C3H) in 150g of water.?
  <<<  propanol is NOT C3H == it is C3H7-OH  <<<  molar mass = 60g/mol
The density of the solution is 1.25g/ml.Calculate:a) molality = = total mass = 151.72g . .  V = m/D = 151.72/1.25  =  121.376 mL  Molarity = (1.72 / 60) / 0.121376     =  0.2361 Mb) mass percent (by mass?)   =  (1.72/151.72) X 100  =  1.134%now you try the other 2