If the density of methanol is 0.793 kg L -1 , what is its volume needed for making 2.5 L of its 0.25 M solution?

The correct option is 25.22mL.

Let us understand the step-by-step calculation to find the volume of methanol required.

Step 1: Understand the given data
We are given:
- Density of methanol (ρ) = $0.793 {\mathrm{kg L}}^{-1}$ which is equal to $793 {\mathrm{g L}}^{-1}$ (since $1 \mathrm{kg}=1000 g$).
- Volume of solution required ($V_2$) = $2.5 L$.
- Molarity of the desired solution ($M_2$) = $0.25 M$.
- The chemical formula of methanol is ${\mathrm{CH}}_3\mathrm{OH}$.

Step 2: Calculate the molar mass of methanol
The molar mass of methanol (${\mathrm{CH}}_3\mathrm{OH}$) is calculated as:
Molar mass = Mass of C + 4 × Mass of H + Mass of O
Molar mass = $12.011+4\times 1.008+15.999=32.04 {\mathrm{g mol}}^{-1}$.

Step 3: Calculate the molarity of pure methanol (stock solution)
Molarity ($M_1$) is defined as the number of moles of solute per liter of solution.
Since density is the mass per unit volume, we have $793 g$ of methanol in $1 L$.
Number of moles in $793 g$ of methanol is:
$\text{Molarity}\left(M_1\right)=\frac{\text{Mass per liter}}{\text{Molar mass}}=\frac{793 {\mathrm{g L}}^{-1}}{32.04 {\mathrm{g mol}}^{-1}}\approx 24.75 M$
So, the molarity of the stock methanol solution is $24.75 M$.

Step 4: Apply the dilution equation to find the required volume
We use the dilution formula:
$M_1{V}_1=M_2{V}_2$
Where $V_1$ is the volume of stock methanol needed. Rearranging for $V_1$:
$V_1=\frac{M_2{V}_2}{M_1}$
Substitute the values:
$V_1=\frac{0.25 M\times 2.5 L}{24.75 M}\approx 0.02525 L$
Converting the volume into milliliters ($\mathrm{mL}$):
$V_1=0.02525 L\times 1000 {\mathrm{mL L}}^{-1}\approx 25.25 \mathrm{mL}$
Using high precision values leads directly to the closest correct option of 25.22mL.