1. How many grams of CaCl2 would be required to produce a 3.5 M (molar) solution with a volume of 2.0 L?

2. What is the molarity of a 5.00 x 102 ml solution containing 2490 g of KI?

3. How many moles of LiF would be required to produce a 2.5 M solution with a volume of 10.5 L?

Question

1.  How many grams of CaCl2 would be required to produce a 3.5 M (molar) solution with a volume of 2.0 L?

2.  What is the molarity of a 5.00 x 102 ml solution containing 2490 g of KI?

3.  How many moles of LiF would be required to produce a 2.5 M solution with a volume of 10.5 L?

anonymous · Answer

1) Ca=40.08 Ci^2 =2(35.45)=70.9
so she did 40.08+70.9=110.98 =1mole of CaCl
3.5m=7/2.0L
7(110.98g)=776.86g

anonymous · Answer

2) K=39.10 l=126.90
39.10+126.90=166.00+166.00=75/5=15m
166.00(15)=2490

anonymous · Answer

3)Li=6.94+F=19.00
6.94+19.00=25.94g
LiF=25.94g
2.5M/10.5L
2.5solution=26.25moles/10.5L

anonymous · Answer

might want to check these ;D