Determine the molal concentration of 71.5 g of linoleic acid,
C12H3202, in 525 g of hexane, C6H14.

0.49 M

Explanation:

Given data:

Mass of solute = 71.5 g

Mass of solvent = 525 g

Molarity of solution = ?

Solution:

Number of moles of solute:

Number of moles = 71.5 g/ 280 g/mol

Number of moles = 0.255 mol

Now we will calculate the molarity:

Molarity =  number of moles of solute/ litters of solution

1 Kg = 1 L

525 g × 1 kg / 1000 g

0.525 kg or 0.525 L

Molarity =  0.255 mol / 0.525 L

Molarity = 0.49 M

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