Question

Given: 36.7 grams of CaF2 is added to 300 mL water. Find molarity?

Answer 1

Answer:

2 M

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Chemistry

Unit 0

• Reading a Periodic Table
• Using Dimensional Analysis

Aqueous Solutions

• Molarity = moles of solute / liters of solution

Explanation:

Step 1: Define

36.7 g CaF₂

300 mL H₂O

Step 2: Identify Conversions

Molar Mass of Ca - 40.08 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol

1000 mL = 1 L

Step 3: Convert

Solute

1. Set up:                               $\displaystyle 36.7 \ g \ CaF_2(\frac{1 \ mol \ CaF_2}{78.08 \ g \ CaF_2})$
2. Multiply:                             $\displaystyle 0.470031 \ mol \ CaF_2$

Solution

1. Set up:                              $\displaystyle 300 \ mL \ H_2O(\frac{1 \ L \ H_2O}{1000 \ mL \ H_2O})$
2. Multiply:                            $\displaystyle 0.3 \ L \ H_2O$

Step 4: Find Molarity

1. Substitute [M]:                    $\displaystyle x \ M = \frac{0.470031 \ mol \ CaF_2}{.3 \ L \ H_2O}$
2. Divide:                                $\displaystyle x = 1.56677 \ M$

Step 5: Check

Follow sig fig rules and round. We are given 1 sig fig as our lowest.

1.56677 M ≈ 2 M