Answer:
[tex]\large \boxed{\text{0.646 mL}}[/tex]
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
MM: 74.12 184.24
CH₃CH₂CH₂CH₂OH + (C₆H₅)₂CHOH ⟶ Product
m/g: 1.30
ρ/g·mL⁻¹: 0.8098
1. Moles of benzhydrol (BH)
[tex]\text{Moles of BH } =\text{1300 mg BH} \times \dfrac{\text{1 mmol BH}}{\text{184.24 mg BH}} =\text{7.056 mmol BH}[/tex]
2. Moles of butan-1-ol (BuOH)
The molar ratio is 1 mmol BuOH:1 mmol BH.
[tex]\text{Moles of BuOH}=\text{7.056 mmol BH} \times \dfrac{\text{1 mmol BuOH}}{ \text{1 mmol BH}} = \text{7.056 mmol BuOH}[/tex]
3. Mass of BuOH
[tex]m = \text{7.056 mmol BuOH} \times \dfrac{\text{74.12 mg BuOH}}{\text{1 mmol BuOH}} = \text{523.0 mg BuOH}[/tex]
4. Volume of BuOH
[tex]V = \text{0.5230 g BuOH} \times \dfrac{\text{1 mL BuOH}}{\text{0.8098 g BuOH}} = \text{0.646 mL BuOH}\\\\\text{You must use $\large \boxed{\textbf{0.646 mL}}$ of butan-1-ol}[/tex]
