As a quality control check, a sample of acetone is taken from a process to determine the concentration of suspended particulate matter. An 850 mL sample was placed in a beaker and evaporated. The remaining suspended solids were determined to have a mass of 0.001 g. The specific gravity of acetone is 0.79 g/cm. (a) Determine the concentration of the sample as mg/L.
(b) Determine the concentration of the sample as ppm (problem from EPA Air Pollution Training Institute)

The volume of the sample given is 850 ml, the density given is 0.79 gram per cm. Now the weight of the sample will be,  

Weight = volume × density = 850 × 0.79  

= 671.5 grams

Weight of the suspended solids given is 0.001 gram

The concentration of the sample can be determined by using the formula,  

Concentration = wt. of sample/volume

= [671.5 - 0.001) 10³ mg / 0.85 L

= 789998.82 mg/L or 789998.82 ppm

Now the concentration of suspended solids is.  

Css = 0.001 × 10³ mg / 0.85 L = 1.1764 mg per L or 1.1764 ppm

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%