<u>Given:</u>
Mass of solvent water = 4.50 kg
Freezing point of the solution = -11 C
Freezing point depression constant = 1.86 C/m
<u>To determine:</u>
Moles of methanol to be added
<u>Explanation:</u>
The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:
ΔTf = kf*m
where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution
ΔTf = 0 C - (-11.0 C) = 11.0 C
m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg
11.0 = 1.86 * moles of methanol/4.50
moles of methanol = 26.613 moles
Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.
We should use renewable resources wisely because <u>if we over use them the resources we already have will decline.</u>
Answer:
Explanation:
The result will be affected.
The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).
Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.
During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.
Thus, NaOH will appear at a higher concentration than it actually is.
Hello friend ☺
ΔH = MCΔT
ΔH = to the amount of energy or change in energy (J)
mass of water
C = waters specific heat capacity
ΔT = change in temperature
and so ΔH = 25 × 4.18 × ( 112-67 ) J = 4702.5 J
Thanks ❤