In the laboratory you dissolve 23.8 g of manganese(II) nitrate in a volumetric flask and add water to a total volume of 125 . mL. What is the molarity of the solution

The correct answer is 1.06 M

Step-by-step explanation:

We have to calculate the molarity (M), which is:

M= moles solute/ 1 L solution.

The chemical formulae of manganese (II) nitrate is Mn(NO₃)₂. So, we first calculate its molecular weight (Mw) as follows:

Mw(Mn(NO₃)₂)= molar mass Mn + (2 x molar mass N) + (6 x molar mass N)= 55 g/mol + (2 x 14 g/mol) + (6 x 16 g/mol) = 179 g/mol

Then, with Mw we calculate the number of moles there is in the given mass of Mn(NO₃)₂:

moles Mn(NO₃)₂= mass/Mw= 23.8 g/(179 g/mol)= 0.133 mol

Now, we need the final volume in liters, so we convert the volume from mL to L:

125 mL x 1 L/1000 mL = 0.125 L

Finally, we divide the moles of Mn(NO₃)₂ into the volume in L, to obtain the molarity in mol/L:

M= 0.133 moles/0.125 L = 1.06 mol/L= 1.06 M