A student weighs out a 6.04 g sample of , transfers it to a 100. mL volumetric flask, adds enough water to dissolve it and then adds water to the 100. mL tick mark. What is the molarity of chromium(III) sulfate in the resulting solution

"0.1540 M" is the correct answer.

Explanation:

The given values are:

Mass of sample,

M = 6.04 g

Volume,

V = 100 mL

or,

  = 100×10⁻³ L

Molar weight of Chromium(III) sulfate

MW = 392.16 g/mol

Now,

The molarity will be:

=

By putting the values, we get

=

=

=

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.