PROBLEMAS DE SOLUCIONES MOLARES

4. Calcular la molaridad de una solución que contiene 2.44 g de Benceno y se encuentra en un volumen de disolución de 340 mL.

C: 12.01x6 = 72.06 uma                               M =  2.44g/78.12g/mol
H:   1.01x6 =   6.06 uma                                               340 mL
                      78.12 g
                                                                     M = 9.186470287x10^-5
                                                                     M = 0.0918 mol/L

5. Calcular la masa de etanol (CH3-CH2-OH) que se encuentra en una solución al 0.48 molar y ocupa un volumen de disolución de 68 mL.

C: 12.01x2 = 24.02 uma                                0.48 mol/L =    X/46.08g/mol    
H:   1.01x6 =   6.06 uma                                                             0.068L
0:       16x1 =      16 uma
                      46.08 g                                    (0.48 mol/L) (0.068 L) =            X             
                                                                                                                 46.08 g/mol
                                                                     
                                                                       X = (0.03264 mol) (46.08 g/mol)
                                                                           = 1.5040512 g
     
SUSTITUCIÓN
  0.48 mol/L =   1.5040512 g/46.08 g/mol  
                                      0.068 L

  0.48 mol/L = 0.48 mol/L