Question

Calculate the molarity of each of the following solutions.

(a) 7.49 g of methanol (CH3OH) in 1.50 × 102 mL of solution:



(b) 9.18 g of calcium chloride (CaCl2) in 2.20 × 102 mL of solution:



(c) 6.51 g of naphthalene (C10H8) in 85.2 mL of benzene solution:

Answer 1

Answer:

a)1.56 M

b) 0.376 M

c) 0.599 M

Explanation:

Step 1: Molarity

Molarity = Number of moles / volume in L

Number of moles = mass / Molar mass

(a) 7.49 g of methanol (CH3OH) in 1.50 * 10² mL of solution

Volume =  1.50 * 10² mL = 0.15 L

Molar mass of CH3OH = 32.04 g/mol

Number of moles CH3OH = 7.49 grams / 32.04 g/mol

Moles of CH3OH = 0.234 moles

Molarity CH3OH = 0.234 moles / 0.15 L

Molarity CH3OH = 1.56 M

(b) 9.18 g of calcium chloride (CaCl2) in 2.20 × 10² mL of solution:

Volume = 2.20 × 10² mL = 0.22L

Molar mass of CaCl2 = 110.98 g/mol

Number of moles CaCl2 = 9.18 grams /110.98 g/mol

Moles CaCl2 = 0.0827 moles

Molarity CaCl2 = 0.0827 moles / 0.22L

Molarity CaCl2 = 0.376 M

(c) 6.51 g of naphthalene (C10H8) in 85.2 mL of benzene solution:

Volume = 85.2 mL = 0.0852L

Molar mass of C10H8 = 128.17 g/mol

Number of moles C10H8 = 6.51 grams / 128.17 g/mol

Moles C10H8 = 0.051 moles

Molarity C10H8 = 0.051 moles / 0.0852 L

Molarity C10H8 = 0.599 M