Answer:
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)= <em>0.133 mol</em>
Now, we need the final volume in liters, so we convert the volume from mL to L:
125 mL x 1 L/1000 mL = <em>0.125 L</em>
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
