Question

A mixture formed of isopropyl alcohol C,H,OH (70%) and water (30%), if you know

that the density of the mixture (at 20°C) = 0.79 g/mL, Calculate :
(1) Molar concentration.
(2) Molal concentration. (C=12 ,H=1 ,O=16 )​

Answer 1

(1) C C3H7OH = 9.200 M

(2) C C3H7OH = 11.647 m

Step-by-step explanation:

mixture:

∴ 70% = (g C3H7OH/g mix)×100

∴ 30% = (gH2O/g mix)×100

∴ δ mix = 0.79 g/mL

assuming:

• g mix = 100g = 0.100 Kg

⇒ V mix = (100g)×(mL/0.79g) = 126.582 mL mix = 0.1266 L mix

⇒ g C3H7OH = 70g

⇒ g H2O = 30g

∴ Mw C3H7OH = 60.1 g/mol

∴ Mw H2O = 18 g/mol

(1) Molar concentration (M):

• M ≡ mol solute/L mix

⇒ C C3H7OH = ((70 g)(mol/60.1 g))/(0.1266L) = 9.200 M

(2) molal concentration (m):

• m ≡ mol solute/Kg mix

⇒ C C3H7OH = ((70 g)(mol/60.1g))/(0.100 Kg) = 11.647 m