Respuesta :
Molarity is moles / L solution. Therefore final concentration is 0.190 moles / L.
A. Converting to g / L
(0.190 moles / L) * (342.296 g/mol) = 65.04 g / L
B. Converting to molality = moles / kg water
(0.190 moles / L) (1 L / 0.955 kg water) = 0.20 molal
C. Converting to mass%
mass% = (0.190 moles / L) * (342.296 g/mol) * (1 mL / 1.02 g) * (1 L / 1000 mL) * 100%
mass% = 6.38%
Answer:
The strength of the sucrose solution is 65.03624 g/L.
Molality of the solution is 0.1989 mol/kg.
Mass percentage of the solution is 6.37%
Explanation:
Strength of the sucrose solution
Molarity of the sucrose solution = 0.190 M = 0.190 mol/L
Molar mass of sucrose = 342.296 g/mol
Strength of the solution = Molarity of the solution × Molar mass of the solute
Strength of the sucrose solution =
= [tex]0.190 mol/L\times 342.296 g/mol=65.03624 g/L[/tex]
Molality of the solution:
[tex]Molality=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}[/tex]
Moles of sucrose in 1L solution = 0.190 moles
Mass of water = 955.0 g = 0.9550 kg
Molality of the solution is:
[tex]Molality=\frac{0.190 mol}{0.9550 kg}=0.1989 mol/kg[/tex]
Mass percentage of the solution:
Density iof the solution =[tex]\rho _{sol'n}=1.02 g/mL[/tex]
Mass of solution = Mass of solute + Mass of water
Volume of solution = 1L = 1,000 mL
[tex]\rho _{sol'n}=1.02 g/mL=\frac{\text{mass of solution}}{\text{Volume of solution}}[/tex]
Mass of solution = 1.02 g/mL × 1,000 mL = 1,020 g
Mass of solute + Mass of water = 1,020 g
Mass of solute = 1,020 g - 955.0 g = 65.0 g
Mass percentage:
[tex]\%(w/w)=\frac{\text{mass of solute}}{\text{mass of solution}}\times 100[/tex]
[tex]\%(w/w)=\frac{65.0 g}{1,020 g}\times 100=6.37\%[/tex]
Mass percentage of the solution is 6.37%
