Question

Volume ofmethanol to make 2L of 0.4M solution
\( (\text{density} = 0.64\text{KgL}^{-1}, \text{ MM} = 32\text{ g mol}^{-1}) \)

Hint:

Always ensure unit consistency. The density is given in Kg/L, but molar mass is usually in g/mol. Convert density to g/L (0.64 Kg/L = 640 g/L) before dividing the mass in grams to prevent major magnitude errors.

Answer 1

Step 1: Understanding the Concept:
To find the required volume of a pure liquid solute to prepare a solution, we must first calculate the required mass of the solute using the molarity equation, and then convert that mass to volume using the given density.
Step 2: Key Formula or Approach:
The required formulas are:
\[ \text{Moles of solute } (n) = \text{Molarity } (M) \times \text{Volume of solution in L } (V) \] \[ \text{Mass } (W) = n \times \text{Molar Mass } (\text{MM}) \] \[ \text{Volume of pure liquid} = \frac{\text{Mass}}{\text{Density}} \] Step 3: Detailed Explanation:
Given:
Molarity (\( M \)) = 0.4 M (mol/L)
Volume of solution (\( V \)) = 2 L
Molar Mass (\( \text{MM} \)) = 32 g/mol
Density = 0.64 Kg/L = 640 g/L (since \( 1 \text{ Kg} = 1000 \text{ g} \)).
First, calculate the required number of moles (\( n \)) of methanol:
\[ n = M \times V = 0.4 \text{ mol/L} \times 2 \text{ L} = 0.8 \text{ moles} \] Next, calculate the mass (\( W \)) of methanol required:
\[ W = n \times \text{MM} = 0.8 \text{ moles} \times 32 \text{ g/mol} = 25.6 \text{ g} \] Now, use the density to find the volume of pure methanol needed:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \] \[ \text{Volume} = \frac{25.6 \text{ g}}{640 \text{ g/L}} \] \[ \text{Volume} = 0.04 \text{ L} \] Convert the volume into milliliters (mL):
\[ 0.04 \text{ L} \times 1000 \text{ mL/L} = 40 \text{ mL} \] Step 4: Final Answer:
The required volume of methanol is 40 mL.