Answer to Question #164462 in General Chemistry for b

Question

What is the % by mass of solute in a 1.43 M solution of NaNO₃ (85.0 g/mol)?  Assume the density of the solution is 1.07 g/cm³.

Answer 1

Let's say the volume is 1 liter, which is equivalent to 1 dm ^ 3 or 1000 cm ^ 3

 Then the mass is 1.07 * 1000 = 1070g - the mass of the solution

 Chemical amount of NaNO3 = 1.43 * 1 = 1.43 mol

 NaNO3 mass = 1.43 * 85.0 = 121.5 g

 Mass fraction of NaNO3 = mass of substance / mass of solution = 121.5 / 1070 = 0.1136 or 11.36%

 In turn, the mass fraction of other components is equal to 1-0.1136 = 0.8864 88.64%