Answer:
Mole fraction of methanol will be closest to 4.
Explanation:
Given, Mass of methanol = 128 g
Molar mass of methanol = 32.04 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given, Mass of water = 108 g
Molar mass of water = 18.0153 g/mol
The formula for the calculation of moles is shown below:
Thus,
So, according to definition of mole fraction:

<u>Mole fraction of methanol will be closest to 4.</u>
Answer:
Option c → Tert-butanol
Explanation:
To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.
The formula is:
ΔT = Kf . m . i
When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.
i = Van't Hoff factor (ions particles that are dissolved in the solution)
At this case, the solute is nonvolatile, so i values 1.
ΔT = Difference between fussion T° of pure solvent - fussion T° of solution.
T° fussion paradichlorobenzene = 56 °C
T° fussion water = 0°
T° fussion tert-butanol = 25°
Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.
Answer: All of the statements are true.
Explanation:
(a) Considering the system mentioned in the equation:-
The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B
(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.
0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,
40 to 60s is 0.14 (0.30-.16 = 0.14) mol
(0.46 > 0.14) mol leading this statement to be true as well.
(c) Average rate from t1 = 40 to t2 = 60 s is given by:
which is true as well
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹