Ask Question
7 September, 19:28

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:

+2
Answers (1)
  1. 7 September, 19:40
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers