Respuesta :
Explanation:
1) Mass of carbon dioxide = 100 g
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide =[tex]\frac{100 g}{44 g/mol}=2.273 moles[/tex]
1 mol = 0.001 kmol
2.273 moles= 2.273 × 0.001 kmol = [tex] 2.273\times 10^{-3} kmol[/tex]
2) 1 liter of ethyl alcohol of density [tex]0.789 g/cm^3[/tex]
Volume of ethyl alcohol ,V= 1 L = 1000 mL
Density of ethyl alcohol =d = [tex]0.789 g/cm^3[/tex]
[tex]1 cm^3=1 mL[/tex]
Mass of ethyl alcohol = m
[tex]m=d\times V=0.789 g/cm^3\times 1000 mL=789 g[/tex]
Molar mass of ethyl alcohol = 46 g/mol
Moles of ethyl alcohol = [tex]\frac{789 g}{46 g/mol}=17.152 mol[/tex]
[tex]17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol[/tex]
3) Volume of oxygen gas,V =[tex]1.5 m^3=1500 L[/tex]
[tex]1 m^3= 1000 L[/tex]
Temperature of the gas = T= 25°C = 298.15 K
Pressure of the gas ,P= 1 atm
Moles of oxygen gas = n
[tex]PV=nRT[/tex]
[tex]n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol[/tex]
0.01632 mol = 0.01632 × 0.001 kmol=[tex]1.632\times 10^{-5} kmol[/tex]
4) Volume water in mixture = 1 L
Density of water = [tex]1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L[/tex]
Mass of water = [tex]1000 g/L\times 1 L = 1000 g[/tex]
Volume of alcohol = 2.5 L
Density of alcohol = [tex]789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L[/tex]
Mass of alcohol = [tex]789 g/L\times 2.5 L = 1972.5 g[/tex]
Mass of mixture = 1000 g + 1972.5 g = 2972.5 g
Mass percentage of water :
[tex]\frac{1000 g}{2972.5 g}\times 100=33.64\%[/tex]
Mass percentage of alcohol :
[tex]\frac{1972.5 g}{2972.5 g}\times 100=66.36\%[/tex]
Moles of water :
[tex]n_1=\frac{1000 g}{18 g/mol}=55.55 mol[/tex]
Moles of alcohol =
[tex]n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol[/tex]
Mole fraction of water :
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644[/tex]
Mole fraction of alcohol :
[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356[/tex]
