Answer to Question #315675 in Inorganic Chemistry for Qybb

Question

Determine the boiling point and freezing point of the following solutions:

5.00g caffeine (C8H10N4O2) and 250.0 ml water.

0.300 kg chloroform (CHCl3) and in 21.0 g eucalyptol (C10H18O).

Answer 1

$\Delta$ T_fr = K_fr * m(molality);

$\Delta$ T_b = E_b * m(molality);

For water: K_fr(H₂O) = 1.86 K * kg/mol;

E_b(H₂O) = 0.52 K * kg/mol;

T_fr(H₂O) = 0.0 C = 273.0 K;

T_b(H₂O) = 100.0 C = 373.0 K;

m(caffeine) = 5.0 g;

M(caffeine) = 194 g/mol;

n(caffeine) = m(caffeine)/M(caffeine) = 5/194 = 0.0258 moles;

V(H₂O) = 250.0 mL;

m(H₂O) = 250.0 g = 0.25 kg;

m(molality of caffeine) = n(caffeine)/m(H₂O) = 0.0258/0.25 = 0.1031 mol/kg

$\Delta$ T_fr(caffeine, water) = K_fr(H₂O) * m(molality of caffeine) = 1.86 * 0.1031 = 0.1918 K;

T_fr(caffeine, water) = T_fr(H₂O) - $\Delta$ T_fr(caffeine, water) = 273 - 0.1918 = 272.8 K = -0.2 C;

$\Delta$ T_b(caffeine, water) = E_b(H₂O) * m(molality of caffeine) = 0.52 * 0.1031 = 0.0536 K;

T_b(caffeine, water) = T_b(H₂O) - $\Delta$ T_b(caffeine, water) = 373 - 0.0536 = 372.95 K = 99.95 C.

For chloroform: K_fr(CHCl₃) = 4.7 K * kg/mol;

E_b(CHCl₃) = 3.6 K * kg/mol;

T_fr(CHCl₃) = -63.4 C = 209.6 K;

T_b(CHCl₃) = 62.0 C = 335.0 K;

m(eucalyptol) = 21.0 g;

M(eucalyptol) = 154 g/mol;

n(eucalyptol) = m(eucalyptol)/M(eucalyptol) = 21/154 = 0.1364 moles;

m(CHCl₃) = 0.300 kg;

m(molality of eucalyptol) = n(eucalyptol)/m(CHCl₃) = 0.1364/0.300 = 0.4545 mol/kg;

$\Delta$ T_fr(eucalyptol, chloroform) = K_fr(CHCl₃) * m(molality of eucalyptol) = 4.7 * 0.4545 = 2.1364 K;

T_fr(eucalyptol, chloroform) = T_fr(CHCl₃) - $\Delta$ T_fr(eucalyptol, chloroform) = 209.6 - 2.1364 = 207.5 K = -65.5 C;

$\Delta$ T_b(eucalyptol, chloroform) = E_b(CHCl₃) * m(molality of eucalyptol) = 3.6 * 0.4545 = 1.6362 K;

T_b(eucalyptol, chloroform) = T_b(CHCl₃) - $\Delta$ T_b(eucalyptol, chloroform) = 335.0 - 1.6362 = 333.4 K = 60.4 C.