Nature">
Documento 179
Documento 179
Documento 179
𝑚 = 2064 𝑔 → 𝑚𝑎𝑠𝑎 𝑑𝑒 𝑙𝑎 𝑙𝑒𝑐ℎ𝑒
12.3 𝑔 1 𝑘𝑔
2064 𝑔 [ ][ ] = 0.253872 𝑘𝑔
100𝑔𝑠𝑜𝑙 1000𝑔
𝑚 𝑇 = 2064 𝑔 → 2.064 𝐾𝑔
𝐾𝑓𝐻2 𝑂 𝑎 0° 𝐶 (𝑅𝑒𝑝𝑜𝑟𝑡𝑎𝑑𝑜 𝑒𝑛 𝑒𝑙 𝑙𝑖𝑏𝑟𝑜 𝐶𝑎𝑠𝑡𝑒𝑙𝑙𝑎𝑛. )
𝐾⋅𝑘𝑔
𝐾𝑓𝐻2 𝑂 = 1.86 𝑚𝑜𝑙
𝜃𝑓 = 𝐾𝑓𝑚
𝑚 = 𝑚𝑜𝑙𝑎𝑙𝑖𝑑𝑎𝑑
𝑚𝑠𝑜𝑙𝑖𝑑𝑜𝑠
𝑚=
𝑀𝑠𝑜𝑙𝑖𝑑𝑜𝑠
𝑚𝑎𝑔𝑢𝑎
𝑝𝑒𝑠𝑜 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟
Despejando M.
𝑚𝑠𝑜𝑙𝑖𝑑𝑜𝑠 𝐾𝑓 𝑚𝑠𝑜𝑙𝑖𝑑𝑜𝑠
𝜃𝑓 = 𝐾𝑓 → 𝑀𝑠𝑜𝑙𝑖𝑑𝑜𝑠 =
𝑚𝑎𝑔𝑢𝑎 𝑀𝑠𝑜𝑙𝑖𝑑𝑜𝑠 𝜃𝑓 𝑚𝑎𝑔𝑢𝑎
Sustituimos:
𝐾 𝐾𝑔
(1.86 ) (0.253872 𝐾𝑔)
𝑀𝑠𝑜𝑙𝑖𝑑𝑜𝑠 = 𝑚𝑜𝑙
(273.15 °𝐾 − 272,6°𝐾)(1.8101 𝐾𝑔)
𝑘𝑔 1000𝑔
𝑀𝑠𝑜𝑙𝑖𝑑𝑜𝑠 = 0.474310 [ ]
𝑚𝑜𝑙 1 𝐾𝑔
𝑀𝑠𝑜𝑙𝑖𝑑𝑜𝑠 = 474.310 𝑔 ⁄𝑚𝑜𝑙
𝜃𝑏 = 𝐾𝑏𝑚 𝑚 = 𝑚𝑜𝑙𝑎𝑙𝑖𝑑𝑎𝑑
𝑚𝑠𝑜𝑙𝑖𝑑𝑜𝑠
𝑚=
𝑚𝑎𝑔𝑢𝑎 𝑃𝑒𝑠𝑜 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟
𝐾𝑘𝑔
Y Kb la constante ebulloscópica del agua del agua a 100°𝐶 = 0.51 𝑚𝑜𝑙
𝐾𝑏 𝑚𝑠𝑜𝑙𝑖𝑑𝑜𝑠
Entonces 𝜃𝑏 = 𝑚𝑎𝑔𝑢𝑎 𝑃𝑀
𝐾𝑘𝑔
(0.51 ) (0.253872 𝐾𝑔)
𝑚𝑜𝑙
𝜃𝑏 =
(1.8101 𝐾𝑔)(0.474310 𝐾𝑔 ⁄ 𝑚𝑜𝑙 )
𝜃𝑏 = 0.150806°𝐾
𝜃𝑏 = 𝑇 − 𝑇°
𝑇° = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑑𝑒 𝑒𝑏𝑢𝑙𝑙𝑖𝑐𝑖𝑜𝑛 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎
𝑇° = 100°𝐶 → 373.15 °𝐾
Temperatura de ebullición:
Hacemos el despeje de T.
𝑇 = 𝜃𝑏 + 𝑇 °
𝑇 = (0.15086°𝐾) + (373.15°𝐾)
𝑇 = 373.3008 °𝐾
El descenso de la presión de vapor:
Δ𝑃 = 𝑋1 𝑃°
𝑃° = 2.5 𝐾𝑝𝑎
Calculamos la X1:
𝐾𝑔
𝑛𝑠𝑜𝑙𝑣𝑒𝑛𝑡𝑒 = 𝑛1 = 1.8101 𝐾𝑔 ⁄ 0.01805 = 100.47738 𝑚𝑜𝑙𝐻2 𝑂
𝑚𝑜𝑙
𝑘𝑔
𝑛𝑠𝑜𝑙𝑖𝑑𝑜𝑠 = 𝑛2 = 0.253872 𝑘𝑔 ⁄ 0.474310 0.53527 𝑚𝑜𝑙𝑠𝑜𝑙𝑖𝑑𝑜𝑠
𝑚𝑜𝑙
0.53524
𝑋1 = = 0.00532701
100.4773
Δ𝑃 = (0.0053271)(2.5 𝐾𝑝𝑎)
Δ𝑃 = 0.013317 𝐾𝑝𝑎
Calcular la presión osmótica:
𝑎𝑡𝑚 𝑙
𝜋 = 𝑅𝑇𝐶 𝑇 = 21.08 °𝐶 → 294.23°𝐾 𝑅 = 0.08205 𝐾 𝑚𝑜𝑙
𝑚 253.872 𝑔 𝑔
𝑃𝑀 = = = 474.28774
𝑛 0.53527 𝑚𝑜𝑙 𝑚𝑜𝑙