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?.

Answer 1

Step-by-step explanation:

1.) Moles of calcium chloride = n

Molarity of the solution = 3.5 M

Volume of the solution =V= 2.0 L

`Molarity=(n)/(V(L))`

`3.5 M=(n)/(2.0 L)`

n = 7 moles

Mass of 7 moles of calcium chloride :

`7 mol* 111 g/mol=777 g`

777 grams of calcium chloride would be required.

2.) Mass of the KI = 2490 g

Moles of KI = n =
`(2490 g)/(166 g/mol)=15 mol`

Volume of the solution,V =
`5* 10^2 ml = 500 mL=0.5 L`

`m=(15 mol)/(0.5 L)=30 mol/L`

Molarity of the KI solution is 30 mol/L.

3.)Molarity of the LiF solution = 2.5 M

Volume of the solution = 10.5 L

Moles of LiF = n

`2.5 M=(n)/(10.5 L)`

n = 26.25 moles

26.25 moles of LiF would be required.

Answer 2

The grams of Cacl2 required to produce a 2 liter solution of 3.5 molarity is determined by multiplying the volume to the molarity, equivalent to 7 moles and multiplying further to the molar mass of CaCl2 which is 110.9 g/mol. The answer is 776.8 grams CaCl2. In 2), molarity is obtained from dividing the moles of solute by the volume of solution. The answer in 2 is 30 molar. In 3) to determine the moles of solute required, we multiply the volume with the given molarity. The answer in 3 hence is 26.25 M.