Acough syrup contains 5.0% ethyl alcohol, c2h5oh, by mass. if the density of the solution is 0.9928 g/ml, determine the molarity of the alcohol in the cough syrup

To answer the question above, let us a basis of the 1000 mL or 1 L. 
                 volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%. 
               mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol). 
                       n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
                       molarity = 1.08 mol/ 1 L = 1.08 M

Idk lolb. jinbbffv nennejejej