CHEM120 FINAL EXAM LATEST QUESTION AND ANSWERS VERSION 2

CHEM120 FINAL EXAM LATEST QUESTION AND ANSWERS VERSION 2
Multiple Choice
1. What is the molarity of a solution containing 55.8 g of MgCl2 dissolved in
1.00 L of solution?
M= moles of solute
liters of solution.
We are given the mass of our solute so we just need to convertit to moles
of MgCl2 via the MM.
55.9 g MgCl2 x 1 mol MgCl2 =0.586111 mols MgCl2
95.2119 g MgCl2
Now we can plug this into our equation for molarityM =
0.587111 mols MgCl2 = 0.586 M MgCl2
`1.00 L
a) 55.8 M
b) 1.71 M
c) 0.586 M
d) 0.558 M
2. What is the mass in grams of Mg(NO3)2 present in 145 mL of a 0.150 Msolution
of Mg(NO3)2?
We know the molarity and the volume, but we need to find the moles.
M= moles of solute
liters of solution. (volume)
First convert mLàL.
145 mLà 0.145 L (divide by 1000)
so we can solve this equation for moles.
Moles= (M) x (Volume)
moles Mg(NO3)2 = (0.150 moles) x (0.145 liters) = 0.02175 molesMg(NO3)2
liter
Now we can convert the number of moles to grams via the molarmass of
Mg(NO3)2.