Respuesta :
Answer: The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %
Explanation:
- Calculating the molarity of solution:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Given mass of camphor = 70.0 g
Molar mass of camphor = 152.2 g/mol
Volume of solution = 575 mL
Putting values in above equation, we get:
[tex]\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M[/tex]
- Calculating the molarity of solution:
To calculate the mass of ethanol, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of ethanol = 0.785 g/mL
Volume of ethanol = 575 mL
Putting values in above equation, we get:
[tex]0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g[/tex]
To calculate the molality of solution, we use the equation:
[tex]\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
where,
[tex]m_{solute}[/tex] = Given mass of solute (camphor) = 70 g
[tex]M_{solute}[/tex] = Molar mass of solute (camphor) = 152.2 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (ethanol) = 451.38 g
Putting values in above equation, we get:
[tex]\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m[/tex]
- Calculating the mole fraction of camphor:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For camphor:
Given mass of camphor = 70 g
Molar mass of camphor = 152.2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol[/tex]
For ethanol:
Given mass of ethanol = 451.38 g
Molar mass of ethanol = 46 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol[/tex]
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
Moles of camphor = 0.459 moles
Total moles = [0.459 + 9.813] = 10.272 moles
Putting values in above equation, we get:
[tex]\chi_{(camphor)}=\frac{0.459}{10.272}=0.045[/tex]\
- Calculating the mass percent of camphor:
To calculate the mass percentage of camphor in solution, we use the equation:
[tex]\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100[/tex]
Mass of camphor = 70 g
Mass of solution = [70 + 451.38] = 521.38 g
Putting values in above equation, we get:
[tex]\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%[/tex]
Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %
