Answer to Question #186808 in Chemistry for MD.Amir Hamza

Question

A solution contains 2P.5 g of cholecalciferol which is also known as Vitamin D3 (C27H44O) in 4500 mL of methanol (CH3OH). [ Here P is the last three digits of your Student ID) [Density of methanol at 25 C is 0.792 g/mL]

(a) Calculate the molarity of this solution

(b) Calculate the mole fraction of Vit D3 and methanol

(c) Is above mixture a dilute or concentrated solution? Explain your understanding within 60 words

Answer 1

(a)

The molarity of the solution equals:

M = n / V = m / (Mr × V)

where M - the molarity, n - the number of moles of solute, V - the volume of solution, m - the mass of solute, Mr - the molar mass (Mr(Vitamin D3) = 384.34 g/mol).

M = m / (Mr × V) = 2.5 g / (384.34 g/mol × 4500 mL) = 2.5 g / (384.34 g/mol × 4.500 L) = 0.0014 M

(b)

The mole fraction equals:

f = n / n_total

where f - the mole fraction, n - number of moles of the substance, n_total - total number of moles.

n(D3) = m(D3)/Mr(D3) = 2.5 g / 384.34 g/mol = 0.0065 mol

n(methanol) = m(methanol) / Mr(methanol) = d(methanol) × V(methanol) / Mr(methanol) = 0.792 g/mL × 4500 mL / 32.04 g/mol = 111.24 mol

f(D3) = 0.0065 mol / (0.0065 mol + 111.24 mol) = 5.84 × 10^-5

f(methanol) = 111.24 mol / (0.0065 mol + 111.24 mol) = 1 mol

(c)

The mixture is a dilute solution because the mole fraction of the solvent (methanol) in the solution is significantly higher than the mole fraction of the solute (vitamin D3).