Amixture 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 )

(1) C C3H7OH = 9.200 M

(2) C C3H7OH = 11.647 m

Explanation:

mixture:

∴ 70% = (g C3H7OH/g mix)×100

∴ 30% = (gH2O/g mix)×100

∴ δ mix = 0.79 g/mL

assuming:

⇒ 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):

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

(2) molal concentration (m):

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

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the answer is c i believe