Answer:
i believe its precipitation??
Explanation:
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
Answer:
I think D
Explanation:
Ok, I'm not sure but it sounds right ish you should check a practice video or something. It might also be B or C but im pretty certain it isnt A just ask yourself is the student measuring it in newtons? Is that important in the process? What about if the student is considering the affect of mass is it important? Good luck srry if im not much of help! If this is like A SUPER IMPORTANT TEST OR SOMETHING RLLLLLLLY IMPORTANT just wait for another answer gl!
The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.
<h3>What is Ideal law ?</h3>
According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.
PV = nRT.
Where,
- p = pressure
- V = volume (1.75 L = 1.75 x 10⁻³ m³)
- T = absolute temperature
- n = number of moles
- R = gas constant, 8.314 J*(mol-K)
Therefore, the number of moles is
n = PV / RT
State 1 :
- T₁ = (25⁰ C = 25+273 = 298 K)
- p₁ = 225 kPa = 225 x 10³ N/m²
State 2 :
- T₂ = 10 C = 283 K
- p₂ = 185 kPa = 185 x 10³ N/m²
The loss in moles of gas from state 1 to state 2 is
Δn = V/R (P₁/T₁ - P₂/T₂ )
V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N
p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)
p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)
Therefore,
Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))
= 0.0213 mol
Hence, The number of moles of gas lost is 0.0213 mol.
Learn more about ideal gas here ;
https://brainly.in/question/641453
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