Answer :
Answer:
"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:
= [tex]\frac{M}{MW\times V}[/tex]
By putting the values, we get
= [tex]\frac{6.04}{392.16\times 100\times 10^{-3}}[/tex]
= [tex]\frac{6.04}{39.216}[/tex]
= [tex]0.1540 \ M[/tex]