Question

9) 13.2 mg of an unknown non-volatile compound is dissolved in
403 mL of water at 25C. If the solution has an osmotic pressure of
10.44 torr, what is the molar mass of the compound? **Show
work**

A) 223 g/mol B) 68.3 g/mol C) 168 g/mol D) 58.3 g/mol E) 195 g/mol

Answer #1

Answer D)58.3 g/mole

= CRT

where

= osmotic pressure of solution, C= concentration of solution (mol/L) and T = temperature of solution in K

Supposw molar mass of given compound = x g/mole

Number of moles of compound = 0.0132 g/ x g/mole =(0.0132/x) moles

Molarity of the solution = Number of moles of solute / Volume of solution in L = (0.0132/x) moles/0.403 L = (0.032754/x) M

= osmotic pressure of solution = 10.44 torr = (10.44/760) atm

T = temperature of solution = (25 +273) = 298 K

Hence

(10.44/760) atm = (0.032754/x) M X 0.082 L.atm/mol.K X 298 K

x = 58.3

molar mass of given compound = 58.3 g/mole

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