A55.50 g gas sample of a gas, at stp, has a volume of 15.0 l. calculate the molecular mass of the gas

We can use the ideal gas law equation to find the number of moles of gas 
PV = nRT
where P - standard pressure - 101 325 Pa
V -volume - 15.0 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - standard temperature  - 273 K
substituting the values in the equation 
101 325 Pa x 15.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 273 K
n = 0.670 mol

molecular mass of the gas can be calculated as follows
molar mass = mass / number of moles 
                     = 55.50 g / 0.670 mol 
 molecular mass of gas is 82.8 g/mol 

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.