Wind blowing sediment against rock or land area, or sandblasting is called a. abrasion c. deflation b. deposits d. picking up PL
2 answers:
You might be interested in
<u>Answer:</u> 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 %
<u>Explanation:</u>
- <u>Calculating the molarity of solution:</u>
To calculate the molarity of solution, we use the equation:
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:
- <u>Calculating the molarity of solution:</u>
To calculate the mass of ethanol, we use the equation:
Density of ethanol = 0.785 g/mL
Volume of ethanol = 575 mL
Putting values in above equation, we get:
To calculate the molality of solution, we use the equation:
where,
= Given mass of solute (camphor) = 70 g
= Molar mass of solute (camphor) = 152.2 g/mol
= Mass of solvent (ethanol) = 451.38 g
Putting values in above equation, we get:
- <u>Calculating the mole fraction of camphor:</u>
To calculate the number of moles, we use the equation:
.....(1)
<u>For camphor:</u>
Given mass of camphor = 70 g
Molar mass of camphor = 152.2 g/mol
Putting values in equation 1, we get:
<u>For ethanol:</u>
Given mass of ethanol = 451.38 g
Molar mass of ethanol = 46 g/mol
Putting values in equation 1, we get:
Mole fraction of a substance is given by:
Moles of camphor = 0.459 moles
Total moles = [0.459 + 9.813] = 10.272 moles
Putting values in above equation, we get:
\
- <u>Calculating the mass percent of camphor:</u>
To calculate the mass percentage of camphor in solution, we use the equation:
Mass of camphor = 70 g
Mass of solution = [70 + 451.38] = 521.38 g
Putting values in above equation, we get:
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 %