Answer to Question #315673 in Inorganic Chemistry for Qybb

Question

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentine. It was first synthesized in 1956 by Nobel prize winner R.B Woodward. It is used as a tranquilizer and sedative. When 1.00g reserpine is dissolved in 25.0g camphor, the freezing point depression is 2.63 0C. Calculate the molality and the molar mass of reserpine.

Answer 1

m(reserpine) = 1 g;

m(camphor) = 25 g;

$\Delta$ T_fr = 2.63 C;

With literature the value K_fr for camphor is 40 °C * kg/mol);

Therefore,

$\Delta$ T_fr = K_fr * m;

m = $\Delta$ T_fr/K_fr = 2.63/40 = 0.06575 mol/kg;

m = n(reserpine)/m(camphor, kg);

n(reserpine) = m * m(camphor, kg) = 0.066 * 0.025 = 0.00164375 moles;

n(reserpine) = m(reserpine)/M(reserpine);

Then,

M(reserpine) = m(reserpine)/n(reserpine) = 1/0.00164375 = 608.4 g/mol.

Answer: the molality (m) = 0.06575 mol/kg;

the molar mass of reserpine (M) = 608.4 g/mol.