# Informe de quimica

Solo disponible en BuenasTareas
• Páginas : 4 (803 palabras )
• Descarga(s) : 0
• Publicado : 22 de agosto de 2012

Vista previa del texto

Determine:

Disolución inicial:

V= 500ml 0.5M x 0.5 L= 0.25 moles

P.M. CuSO4 . H2O = 249.69 g/mol 1L

M = 0.5

%p/p= 62.4g x 100 =12.48%p/p %p/v = 62.4g x 100 = 12.48%p/v

500 g 500ml

P.M. CuSO4 . H2O = 249.69 g/mol nequiv. = 62.4g =0.49 equiv.

PE= PM / valencia del catión 125g/eq.

PE= 249.69 /2

PE= 125 g/eq. N = 0.49 equiv = 0.98equiv/L

0.5 L

▪ Disolución A ( 0.2 M)

Volumen extraído: 0.5M x Vi = 0.2M x 50mlVi = 0.2M x 50 ml Vi = 20ml

0.5 M

Moles extraídos: 0.025 moles

0.2M x 0.05 L = 0.01 moles0.01M x 249.69 g/mol = 2.5 g

1L 1M

| |masa |volumen|densidad |Nomoles |P.M. |
|soluto |2.5 g |2.49 ml | |0.01 moles |249.69 g/mol|
|solvente |47.5 g |47.5 ml |1 g/ml |2.63 moles |18 g/mol |
|solución |50 g|50 ml |1 g/ml |2.64 moles | |

%p/p = 2.5 g x 100 = 5 %p/p %p/v = 2.5g x 100 = 5 %p/v50 g 50 ml

Xsoluto = 0.01 moles = 0.00378 M = 0.01 = 0.2 M

2.64 moles...