Question

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

Answer 1

molarity = number of mole of solute/volume of solvent in liters
0.25 = number of mole of methanol/2.5
no. of mole of methanol = 0.25×2.5 = 0.625 mole
weight of methanol = no. of mole of methanol×molecular weight of methanol
                            = 0.625×32
                            = 20 gram
weight of methanol = 20/1000 kg
volume = weight of methanol/density
           = (20/1000)/0.793
           = 20/793
           = 0.025liter

Answer 2

25.2 mL is Volume Needed

Explanation:

Moles of methanol present in 2.5 L of 0.25 M solution

$Molarity = \frac{Moles of CH3OH}{Volume in litre}$

$0.25 = \frac{Moles of CH3OH}{2.5}$

Moles of CH3OH = 2.5 * 0.25 = 0.625

Mass of CH3OH = 0.625 * 32 = 20 g

(∵ Molecular mass of CH3OH = 12 + 1 * 3 + 16 + 1 = 32)

0.793 * $10^{3}$ g of CH3OH is present in 1000 mL

20 g of CH3OH is present in

= $\frac{1000}{0.793 * 10^{3} } * 20 =25.2$ mL