8.03 solutions report is described below.
Explanation:
8.03 Solutions Lab Report
In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.
Pre-lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
The correct answer is base
Exothermic is the right answer my friend
Answer:
The solution is not ideal.
The relative strengths of the solute-solvent interactions are greater compared to the solute-solute and solvent-solvent interactions
Explanation:
The total vapor pressure is the sum of the partial pressures of water and methanol, and they are calculated by the Raoult´s law equation:
Pₐ = Xₐ Pºₐ, where Pₐ is the partial pressure of component A
Xₐ is the molar fraction of A
P⁰ₐ is the pressure of pure A
So lets calculate the partial pressures of methanol and water and compare them with the given total vapor pressure of solution:
X H2O = 0.312 ⇒ X CH3OH = 1 - 0.312 = 0.688
PH2O = 0.312 x 55.3 torr = 17.3 torr
PCH3OH = 0.688 x 256 torr = 176.1 torr
Ptotal = PH2O + PCH3OH = 17.3 torr + 176.1 torr = 193.4 torr
This pressure is less than the experimental value of 211 torr. So the solution is not ideal. The relative strength of the solute-solvent interactions are greater than the solute-solute and solvent-solvent interactions.
The reason for this is the presence of hydrogen bonding between methanol and water.
Here, we apply a mass balance:
Moles of chloride ions in final solution = sum of moles of chloride ions in added solutions
We must also not that each mole of sodium chloride will release one mole of chloride ions, while each mole of magnesium chloride will release two moles of chloride ions.
Moles = concentration * volume
Moles in final solution = moles in NaCl solution + moles in MgCl₂ solution
C * (150 + 250) = 1.5 * 150 + 2 * 0.75 * 250
C = 1.5 M
The final concentration is 1.5 M
