ROUND ALL TO 2 DECIMAL PLACES Question 4
What is the molarity of the following solution?
1.71 mol in 5.15 L of solution

Question 5
How many moles of solute are contained in the following solution?
126mL of 4.03 M CaCl2

Question 6
How many grams of solute are contained in the following?
102.7 ML of 7.99 M KOH

Explanation:

4) What is the molarity of the following solution?

Number of moles = 1.71 mol

Volume of solution = 5.15 L

Molarity of solution = ?

Solution:

Molarity = number of moles / volume in L

Molarity = 1.71 mol / 5.15 L

Molarity = 0.33 M

5) How many moles of solute are contained in the following solution?

Volume of solution = 126mL  = 126 mL× 1 L /1000 mL = 0.126 L

Molarity of solution = 4.03 M

Number of moles of CaCl2 = ?

Solution:

M = number of moles / volume in L

4.03 M  =  number of moles /  0.126 L

Number of moles = 4.03 M  × 0.126 L ( M = mol / L)

Number of moles = 0.508 mol

6) How many grams of solute are contained in the following?

Volume of solution = 102.7 mL (102.7 mL× 1 L /1000 mL = 0.1027 L)

Molarity  = 7.99 M

Mass of solute (KOH) = ?

Solution:

M = number of moles / volume in L

7.99 M = number of moles / 0.1027 L

Number of moles = 7.99 M × 0.1027 L

Number of moles = 0.82 mol

Mass of KOH:

Mass = number of moles × molar mass

Mass = 0.82 mol × 56.1 g/mol

Mass = 46.00 g