A student dissolves 20. g of benzoic acid (C,H,0,) in 125. mL of a solvent with a density of 1.07 g/mL. The student notices that the volume of the solvent does not change when the benzoic acid dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits. molarity = molality %3D olo

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### Problem Statement

A student dissolves 20.0 g of benzoic acid (C\(_7\)H\(_6\)O\(_2\)) in 125. mL of a solvent with a density of 1.07 g/mL. The student notices that the volume of the solvent does not change when the benzoic acid dissolves in it.

**Task:** Calculate the molarity and molality of the student's solution. Ensure each of your answer entries has the correct number of significant digits.

![Graphical representation to input molarity and molality with units](The image shows placeholders to enter the calculated molarity and molality values, with the molarity indicated in units of molarity and molality in units of molality.)

### Details

1. **Given Data:**
   - **Mass of benzoic acid:** 20.0 g
   - **Volume of solvent:** 125. mL
   - **Density of solvent:** 1.07 g/mL

2. **Molar Mass Calculation:**
   - **Molar mass of benzoic acid (C\(_7\)H\(_6\)O\(_2\)):**
     - Carbon (C): 12.01 g/mol
     - Hydrogen (H): 1.01 g/mol
     - Oxygen (O): 16.00 g/mol
   
   Formula: C\(_7\)H\(_6\)O\(_2\)
   - Molar mass = (7 × 12.01) + (6 × 1.01) + (2 × 16.00)
   - Molar mass = 84.07 + 6.06 + 32.00 = 122.13 g/mol

3. **Molarity Calculation:**
   - **Volume of solution in liters:** 125 mL = 0.125 L
   - **Moles of benzoic acid:**
     - Moles = Mass / Molar mass = 20.0 g / 122.13 g/mol = 0.164 mol
   - **Molarity (M):**
     - Molarity = Moles / Volume = 0.164 mol / 0.125 L = 1.31 M

4. **Molality Calculation:**
   - **Mass of solvent in grams:**
Transcribed Image Text:### Problem Statement A student dissolves 20.0 g of benzoic acid (C\(_7\)H\(_6\)O\(_2\)) in 125. mL of a solvent with a density of 1.07 g/mL. The student notices that the volume of the solvent does not change when the benzoic acid dissolves in it. **Task:** Calculate the molarity and molality of the student's solution. Ensure each of your answer entries has the correct number of significant digits. ![Graphical representation to input molarity and molality with units](The image shows placeholders to enter the calculated molarity and molality values, with the molarity indicated in units of molarity and molality in units of molality.) ### Details 1. **Given Data:** - **Mass of benzoic acid:** 20.0 g - **Volume of solvent:** 125. mL - **Density of solvent:** 1.07 g/mL 2. **Molar Mass Calculation:** - **Molar mass of benzoic acid (C\(_7\)H\(_6\)O\(_2\)):** - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol - Oxygen (O): 16.00 g/mol Formula: C\(_7\)H\(_6\)O\(_2\) - Molar mass = (7 × 12.01) + (6 × 1.01) + (2 × 16.00) - Molar mass = 84.07 + 6.06 + 32.00 = 122.13 g/mol 3. **Molarity Calculation:** - **Volume of solution in liters:** 125 mL = 0.125 L - **Moles of benzoic acid:** - Moles = Mass / Molar mass = 20.0 g / 122.13 g/mol = 0.164 mol - **Molarity (M):** - Molarity = Moles / Volume = 0.164 mol / 0.125 L = 1.31 M 4. **Molality Calculation:** - **Mass of solvent in grams:**
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