Reporte

Solo disponible en BuenasTareas
  • Páginas : 2 (291 palabras )
  • Descarga(s) : 0
  • Publicado : 15 de febrero de 2011
Leer documento completo
Vista previa del texto
Tipo de solución | Concentración Molar | Valor de pH |
Café diluidoCafé=1.2 gAzúcar=28.6gAgua=250ml | Cafeína M= m/pm C8H10N4O2l8x12.01=98.08 = 1.2g/194.06g/mol10x1.00=10.00 0.25 l4x14.00=56 =0.0247mol/l2x 15.99=31.98 Total 194.06g/molSacarosa= 28.06g/342.01g/molC12H22O11 0.25 l12x12.01=144.12 =0 .3281 mol/l22x1.00=22.0011x15.99=175.89Total 342.01g/mol | pH= ( H+ )+ ( H- )pH= 0.0247 + 0.3281pH=0.0628pH=-Log(0.0628)pH:1.2020 |
Agua LimónAgua= 200mlAzúcar=21.5Limón=25mlDensidad del acido Cítrico = 1.665g/ml(1.665g/ml)(25ml)=41.625 g | SacarosaM= m/pm C12H22O11 l12x12.01=144.12 21.05g/342.01g/mol 22x1.00=22.00 0.20 l11x15.99=175.89=0.3077 mol/lTotal 342.01g/molAcido Cítrico = 41.625g/191g/mol C6 H8O7 0.20l6x12.01=72.06=1.0896mol/l8x1.00=8 , 7x15.99=111.93 | pH= ( H+ ) + ( H- )pH= 0.3077 + 1.0896pH= 1.3973pH=-Log(1.3973)pH:-0.1452 |
Vinagre50mlDensidad del vinagre: 1.0056 g/cm31000cm³=1litro 50cm3 = 0.05litroMasa=(1.0056g/cm3)(50cm3)=50.28g | CH3 COOH = 60g/mol M= m/pm l M=50.28g/60g/mol 0.05l= 16.76mol/l | pH=-Log(16.76)pH= -1.2242 |
Vinagre conAguaVinagre: 50 mlAgua: 50ml1000cm3 = 1 litro 50cm3 = 0.05 litro1gramo por Cm350cm3 x 1g/cm3=50g | CH3 COOH = 60g/molM=50.28g/60g/mol 0.05l= 16.76mol/lH2O2x1.00=2.001x15.99=15.99Total: 17.99g/mol M= m/pm l = 50g/17.99g/mol 0.05 l =55.5864mol/l | pH= ( H+ ) + ( H- )pH= (16.76) + (55.5864)pH=72.3464pH= -Log(72.3464)pH= -1.8594 |
tracking img