Answer to Question #295241 in Inorganic Chemistry for Jamael

Question

Five hundred (500) milliliters of aqueous solution contains thirty (30) milligrams of zinc nitrate (Zn(N03)2 . The density of the aqueous solution is 1 gram per milliliter (1 g/ml). The atomic masses : Zn = 65.39 , N = 14.00674 , O = 15.994 , Compute for the concentration of zinc nitrate in the aqueous solution in terms of mass fraction (mi), parts per million (ppm), parts per billion (ppb), parts per trillion (ppt), milligrams per liter (m/l) and Molarity (M).

Answer 1

Mass fraction $=\frac{Mass\>of \>Zinc\>Nitrate}{Total\>mass}$

$=\frac{30×10^{-3}}{500×1}$

$=0.00006$

Parts per million$=0.00006×10^6$

$=60ppm$

Parts per billion $=60000ppb$

Parts per trillion $=6×10^7ppt$

Milligram per litre $=\frac{30}{500}×1000$

$=60m/L$

Molar mass of Zn(NO₃ )₂ $=65.39+2[14.00674+3(15.994)]$

$=189.36748g$

Molarity $=\frac{60×10^{-3}}{189.36748}$

$=0.000316844M$