Química organica CÁLCULOS

CÁLCULOS

- Calcular cantidad necesaria (g o L) para preparar soluciones de 50 mL 0.1 M.

- CH3CH2OH(l) 0.1 M, d= 0.79 g/cm3 
P.M.= (2mol)(12g) + (6mol)(1g) + (1mol)(16g)= 46 g/mol
g=P.M*V*M = (46 g/mol)(.05L)(0.1mol/L)= 0.23 g
v= m/d = (0.23g)/(.79g/cm3)= .2911 cm3 de CH3CH2OH  

- NaCl(s) 0.1M 99% pureza
P.M.= (1mol)(23g) + (1mol)(35.45g)= 58.45 g/mol
g=P.M*V*M = (58.45 g/mol)(.05L)(0.1mol/L)= .2922 g de NaCl
100g --- 99g                  X= (100g)(0.2922g) / 99g = 0.2951 g de NaCl
X ------ .2922 g


- CuCl2(s)  0.1 M
P.M.= (1mol)(63.55g) + (2mol)(35.45g)= 134.45 g/mol
g=P.M*V*M = (134.45g/mol)(.05L)(0.1mol/L)= 0.67225 g de CuCl2

- KOH(s) 0.1 M 85% pureza
P.M.= (1mol)(39g) + (1mol)(16g) + (1mol)(1g) = 56 g/mol
g=P.M*V*M = (56 g/mol)(.05L)(0.1mol/L)= 0.28 g
100g --- 85g                  X= (100g)(0.28g) / 85g = 0.3294 g de KOH
X ------ 0.28 g

- H2SO4(l) 0.1 M
P.M.= (2mol)(1g) + (1mol)(32g) + (4mol)(16g)= 98 g/mol
g=P.M*V*M = (98 g/mol)(.05L)(0.1mol/L)= .49 g de H2SO4(l)

- HCl(l) 0.1 M, 37% pureza, d= 1.170 g/cm3 
P.M.= (1g)(1mol) + (35.45g)(1mol)= 36.45 g/mol
g=P.M*V*M = (36.45 g/mol)(.05L)(0.1mol/L)= 0.18225 g
100g --- 37g                  X= (100g)(0.18225g) / 37g = 0.4925g
X ------ 0.18225 g
v= m/d = (0.4925g)/(1.170g/cm3)= 0.4209 cm3 de HCl

- NaOH(s)  0.1M, 97.7%pureza
P.M.= (1mol)(23g) + (1mol)(16g) + (1mol)(1g)= 40g/mol
g=P.M*V*M = (40 g/mol)(.05L)(0.1mol/L)= 0.2 g
100g --- 97.7g                  X= (100g)(0.2g) / 97.7g = .2047 g NaOH
X ------ 0.2 g

- HNO3(l) 0.1 M
P.M.= (1mol)(1g) + (1mol)(14g) + (3mol)(16g)= 63 g/mol
g=P.M*V*M = (63 g/mol)(.05L)(0.1mol/L)= 0.315 g de HNO3(l)

- CH3COOH 0.1 M
P.M.= (2mol)(12g) + (4mol)(1g) + (2mol)(16g)= 60 g/mol
g=P.M*V*M = (60 g/mol)(.05L)(0.1mol/L)= 0.3 g de CH3COOH

- C12H22O11 0.1 M
P.M.= (12mol)(12g) + (22mol)(1g) + (11mol)(16g)= 342 g/mol
g=P.M*V*M = (342 g/mol)(.05L)(0.1mol/L)= 1.71 g de C12H22O11

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