A solution was prepared by dissolving 476 mg of silver chromate (Ag2CrO4, MW = 331.73 g mol-1 ) in 633 mL of water a) moles of Ag2CrO4 b) millimoles of Ag2CrO4 c) molarity of Ag2CrO4 d) molarity of Ag+ e) molarity of CrO4 2- f) parts per million of Ag2CrO4 g) % (weight/volume) Ag2CrO4 (in units of g/mL)

Silver chromate, Ag2CrO4 , a sparingly soluble compound, has  a solubility product, Ksp , of 1.2 x 10-12

What is the molar  solubility of Ag2CrO4 in pure water?

Explanation:

Use the ICE approach:       Write the Ksp equilibrium:

     Ag2CrO4(s) = 2 Ag+(aq) + CrO4  2- (aq)            Ksp = 1.2 x 10 -12

I :           1         0         0

C :      (“0”)    +2x      +x

E :          1        2x        x

Note: above, we let          x = [CrO4   2- ] and so [Ag+] = 2x )

Ksp =   [Ag+] 2 [CrO4   2-]/1 = (2x)2(x) = 4x3 = 1.2 x 10-12

Or,        x3  = 3.0 x 10 -13        Take cube root of both sides:

        => x = 6.7 x 10 -5 M = [ CrO4   2- ] => [ Ag+ ] =2x = 1.3 x 10  -4 M

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

fe 55.85 g/mol and h2o 18.02 g/mol.