Answer :
The molarity of the prepared solution : 0.014 M
Further explanation
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
[tex]\tt \large {\boxed {\bold {M ~ = ~ \dfrac {n} {V}}}[/tex]
Where
M = Molarity
n = number of moles of solute
V = Volume of solution
mass of solute (Co(NO₃)₂) :253 mg = 0.253 g
moles of solute (Co(NO₃)₂) :
[tex]\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.253~g}{182.94}\\\\mol=0.0014[/tex]
Volume of solution = 100 ml=0.1 L
The molarity :
[tex]\tt molarity(M)=\dfrac{0.0014}{0.1}=0.014[/tex]