Question

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

Answer 1

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

Step-by-step 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((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((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 = (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