alguien me puede ayudar en esto ES PARA EL MIERCOLES X FA
Respuestas a la pregunta
1.Mm Ca(OH)2 = 74,09 g/mol
Calcular moles ( n= m/Mm)
n = 45 g / 74.09 g/mol = 0.607 moles
v = 250 / 1000 = 0.250 L
M = 0.607 mol/ 0.250 L = 2.42
2.Mm C12H22O11 = 342 g/mol
Calcular moles
n = 150 g / 342 g/mol = 0.438 moles
masa disolvente = 300 / 1000 = 0.300 Kg de agua
m = 0.438 mol / 0.300 Kg = 1.46
3. Mm C6H12O6 = 180 g/mol
Calcular moles ( n= m/Mm)
n = 80 g / 180 g/mol = 0.444 moles
v = 300 / 1000 = 0.30 L
M = 0.444 mol/ 0.300 L = 1.48
4. Calcular moles ( n= m/Mm)
n = 120 g / 60 g/mol = 2 moles
v = 750 / 1000 = 0.750 L
M = 2 mol/ 0.750 L = 2.67
5. Cuál será la molaridad de una solución que contiene 20 gramos de cloruro de sodio NaCl en 5 litros de solución?
Calcular moles ( n= m/Mm)
n = 20 g / 58.5 g/mol = 0.341 moles
M = 0.341 mol / 5 L = 0.0682
6. Calcular moles ( n= m/Mm)
n = 50 g / 60 g/mol = 0.833 moles
v = 850 / 1000 = 0.850 L
M = 0.833 mol/ 0.850 L = 0.98
7.Mm Cu(SO4) = 159,60 g/mol
Calcular moles (M = n/V(L)
n = 2.0 mol/L x 1 L = 2.0 mol
calcular gramos (n = m/Mm)
masa = 2.0 mol x 159.60 g/mol = 319.2 g de sulfato cuprico
8. Mm K2Cr2O4 = 194 g/mol
Calcular moles ( n= m/Mm)
n = 25.0 g / 194 g/mol = 0.128 moles
v = 300 / 1000 = 0.300 L
M = 0.128 mol/ 0.300 L= 0.42
9. Mm HCl = 36.5 g/mol
Calcular moles
n = 441 g / 36.5 g/mol = 12.082 moles
masa disolvente = 1500/ 1000 = 1.5 Kg de agua
m = 12.082 mol / 1.5 Kg = 8.05
10. Mm C2H5OH = 46 g/mol
Calcular moles ( m= n/Kg)
n = 1.54 mol/Kg x 2.5 Kg = 3.85 moles
calcular gramos ( n= m/Mm)
masa = 3.85 mol x 46 g/mol = 177.1 g
11. Mm CaCO3 = 100 g/mol
V = 250 / 1000 = 0.250 L
Calcular moles (M = n / V(L)
n = 2.0 mol/L x 0.250 L = 0.500 mol
calcular gramos (n = m/Mm)
masa = 0.5 mol x 100 g/mol
masa = 50 g de CaCO3
12. Mm C12H22O11 = 342 g/mol
Calcular moles
n = 20.0 g / 342 g/mol = 0.058 moles
masa disolvente = 125 / 1000 = 0.125 Kg de agua
m = 0.058 mol / 0.125 Kg
m = 0.46
13. Mm NaCl = 58.5 g/mol
V = 250 / 1000 = 0.250 L
Calcular moles (M = n / V(L)
n = 2.5 mol/L x 0.250 L = 0.625 mol
calcular gramos (n = m/Mm)
masa = 0.625 mol x 58.5 g/mol
masa = 36.56 g de HCl
14. Mm Na2SO4 = 142 g/mol
Calcular moles
n = 500 g / 142 g/mol = 3.52 mol
calcular volumen ( M = n/V(L) )
V(L) = 3.52 mol / 0.75 mol/L
V (l) = 4.69
15. Mm CaCl2 = 110,98 g/mol
Masa disolvente = 1500 / 1000 = 1.50 Kg de agua
Calcular moles
n = 250 g / 110.98 g/mol = 2.252 mol
calcular molalidad
m = 2.252 mol / 1.50 Kg = 1.50
16. Mm Na2CO3 = 106 g/mol
V = 100 / 1000 = 0.100 L
Calcular moles
n = 50 g / 106 g/mol = 0.471 mol
calcular Molaridad
M = 0.471 mol / 0.1 L = 4.71
calcular normalidad
Peq. = Mm/Valencia
Peq. = 106 / 2 = 53 g/eq.
Calcular nº eq.g = masa/ Peq.
Nº eq-g. = 50 g / 53 g / eq-g = 0.94
N = 0.94 eq-g / 0.1 L = 9.4
17. Mm = 43 g/mol
Peq-g = Be(OH)2 = 43 /2 = 21.5
calcular normalidad
Peq. = Mm/Valencia
Calcular nº eq.g = masa/ Peq.
Nº eq-g. = 20 g / 21.5 g / eq-g = 0.93
N = 0.93 eq-g / 0.7 L = 1.33
18. Peq = 40 /1 = 40 g/eq
V = 0.500 L
Calcular nº eq-g =
N = nº eq-g/ V(L)
Nº Eq-g = 4.5 eq-g/L X 0.500 L = 2.25
Nº eq-g = masa/Peq.
Masa = 2.25 eg x 40 g/eq. = 90 g de NaOH
19. Mm HCl = 36.5 g/mol
Peq. = Mm/Nº H
Peq. = 36.5 /1 0 36.5 g/eq.
Calcular nº eq.g = masa/Peq.
Nº eq-g = 100 g / 36.5 g/ eq-g = 2.74
N = 2.74 eq-g/ 3 L = 0.91
20. Calcular moles totales
nt = moles totales de la disolución = nmetanol + netanol + netanol = 10 + 1+ 8 = 19
xmetanol = nmetanol / nt = 10 / 19 = 0,53
xetanol = netanol / nt = 1 / 19 = 0,05
xagua = netanol / nt = 8 / 19 = 0,42
comprobar que la solución es igual a 1:
xmetanol + xetanol + xagua = 0,53 + 0,05 + 0,42 = 1
21. Mm C2H5OH = 46 g / mol
Mm H2O = 18 g / mol
moles C2H5OH = n C2H5OH. = 40 g / 46 g / mol = 0,87 moles
moles H2O = n H2O = 100 g / 18 g / mol = 5,56 moles
moles totales disolución = nt = 0,87 + 5,56 = 6,43 moles
X C2H5OH = n C2H5OH / nt = 0,87 / 6,43 = 0,14
XC2H5OH = X H2O = n H2O / nt = 5,56 / 6,43 = 0,86
XC2H5OH + X H2O = 0,14 + 0,86 = 1
22. Mm CH3OH = 32 g / mol
Mm H2O = 18 g / mol
moles de C3H8O3: n glicerina = 70 / 92 = 0,76 moles
moles de CH3OH: n metanol = 20 / 32 = 0,62 moles
moles de H2O: n agua = 250 / 18 = 13,89 moles
moles totales: n total = = 15,27 moles
glicerina: X C3H8O3 = n C3H8O3 / ntotal = 0,76 / 15,27 = 0,050
etanol: X CH3OH = n CH3OH / n total = 0,62 / 15,27 = 0,041
agua: X H2O = n H2O / n total = 13,89 / 15,27 = 0,910
suma de fracciones molares es igual a 1: 0,05 + 0,041 + 0,910 = 1
23. X H2O: X H2O = 0,56 / n total = 0,9 entonces n total = 0,56 / 0,9 = 0,62
X CH3OH = n CH3OH/ n total = 0,056 = n CH3OH = 0,056 x 0,62 = 0,035
X C3H8O3 = n C3H8O3 / n total = 0,044 = n C3H8O3 = 0,044 x 0,62 = 0,027
n H2O = 0,56 = masa agua / Mm H2O = masa H2O = 0,56 x 18 = 10,1 g
n CH3OH = 0,035 = masa CH3OH / Mm = masa CH3OH = 0,035 x 32 = 1,12 g
n C3H8O3 = 0,027 = masa C3H8O3/ Mm = masa glicerina = 0,027 x 92 = 2,48 g