Explanation:
impurities affect both melting point and boiling point
in the case of melting point it lowers the melting point
so I guess one of the samples Ie contained impurities
Diffusion is the process of a substance spreading out to evenly fill its container or environment. Rate of diffusion of a gas is inversely proportional to the molar mass of the gas.

Lighter(lower) the molar mass of the gas , faster will be its rate of diffusion and heavier (higher) the molar mass of the gas , slower will be its rate of diffusion.
We have to arrange the given gases from slowest rate of diffusion to fastest rate of diffusion that means we need to arrange gases from higher molar mass to lower molar mass.
Molar mass of given gases are :
Cl = 35.5 g/mol
Xe = 131.29 g/mol
He = 4.00 g/mol
N = 14.00 g/mol
So correct order for slowest rate of diffusion (highest molar mass) to fastest rate of diffusion (lowest molar mass) is :
Xe , Cl , N , He
Xe having the highest molar mass will have the slowest rate of diffusion and He with lowest molar mass will have the fastest rate of diffusion, so option 'c' is correct.
Note : Slowest rate of diffusion = High Molar Mass
Fastest rate of diffusion = Low Molar Mass
The motivation to abstain from adding water to concentrated acids is that, with a few acids, amid weakening, a considerable measure of warmth is discharged, by adding the corrosive to the water, the generally extensive measure of water will retain the warmth. On the off chance that you added water to concentrated corrosive when you initially beginning pouring the water, it could get sufficiently hot for the little measure of water that was filled all of a sudden bubble and splatter corrosive on you. Concentrated sulfuric corrosive is most famous for doing this, not all acids get that hot on weakening, but rather in the event that you make a propensity for continually adding the corrosive to water for every one of them, you can't turn out badly.
The student was not successful.
Consider the standard reduction potentials.
Li⁺ + e⁻ ⇌ Li; E° = -3.04 V
2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻; E° = -0.83 V
To reduce Li⁺ to Li, the student must apply 3.04 V.
However, it takes only 0.83 V to reduce water to hydrogen.
Thus, the student will get H₂ instead of Li.