Specific gravity of a 33% (NH4)SO4 solution is 1.1 what mass of (NH4)SO4 would be required to prepare 201mL?

Question

the end of the ? is 201 ml "of this soluition".
It was over 100 characters so it couldnt fit but yeah please help =]

Answer 1

You're not specific if the 33% solution is w/v or w/w, most likely w/w.
from the definition of density (or specific gravity) d=m/V. specific gravity is defined as density of object/density of water, so numerically specific gravity equals denisity.
so: m=d*V=1.1 g/cm3*201 cm3=221.1g - this is the mass of solute and water combined. Percent concentration by definition is C%=mass of solute*100%/mass of solution, so mass of solute=C%*mass of solution/100%=33%*221.1g/100%=72.96g

Answer 2

201 ml of the soln weighs = 1.1x201 = 221.1 g

Mass of (NH4)2SO4 = .33 x 221.1 = 72.963 g