Polar

Anexo de cálculos:

Para calcular las diluciones:

Cc= 20g/ 50ml = 0.4g/ml

100% ________ 0.4g/ml
80% _________x= 0.32g/ml

1ml __________0.32g soluto
50ml _________x= 16g soluto

0.4g _____ 1ml _______ 100%
16g _____x= 40ml solución madre


100% ________ 0.4g/ml
60% _________x=0.24g/ml

1ml __________ 0.24g soluto
50ml _________x= 12g soluto

0.32g _____ 1ml _______ 80%
12g _____x= 37.5ml solución madre


100% ________0.4g/ml
40% _________x= 0.16g/ml

1ml __________ 0.16g soluto
50ml _________x= 8g soluto

0.24g _____ 1ml _______ 60%
8g _____x= 33.33ml soluciónmadre


100% ________ 0.4g/ml
20% _________x= 0.08g/ml

1ml __________ 0.08g soluto
50ml _________x= 4g soluto

0.16g _____ 1ml _______ 40%4g _____x= 25ml solución madre

Volumen necesario para el llenado del tubo:
C1 x V1=C2 x V2

80% ___ 100%. X = 80%. 35ml => X = 28ml
60% ___ 80%.X = 60%. 35ml => X = 26.25ml
40% ___ 60%. X = 40%. 35ml => X = 23.33ml
20% ___ 40%. X = 20%. 35ml => X = 17.5ml

Cálculo del poder rotatorioespecífico:

α = [α].l.cc [α] = m/l l = 2.10dm m= α/cc

[α] = m/l = 128.93ºg/ml / 2.10dm = 61.38095238ºml/g.dm

Δ[α]/[α] = Δm/m + Δl/l
=> Δ[α]= [0.01ºg/ml / 128.93ºg/ml + 0.01dm / 2.10dm] x 61.38095238ºml/g.dm
Δ[α] = 0.297052154ºml/g.dm

Pureza:

Experimental: 128.93

Teórica: m=66.37ºml/g.dm x 2.10dm = 139.38ºml/g

139.38ºml/g_____100%
128.93ºml/g______ x= 92.48%

92.48%_______61.38ºml/g.dm
100%________ x= 66.37ºml/g.dm