Based on the freezing point depression value of HCl in benzene, HCl does not act as an electrolyte in benzene.
<h3>What is the effect of solute on the freezing point of a solvent?</h3>
The freezing point of a solvent such as water is lowered by the solute which ionizes in it.
The freezing point of a solution depends on the number of particles in the solution.
The formula for freezing point depression is given below:
ΔTf = i * Kf * m
HCl remains as molecules and does not ionize in a solution of Benzene. This is because benzene is a nonpolar solvent.
Therefore, the equation for the dissolution of HCl in benzene is given below:
HCl (g) → HCl (in benzene solution)
1 mol of HCl gas gives 1 mol of HCl molecules in solution.
Therefore, HCl remains a molecule and does not act as an electrolyte in benzene.
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Explanation:
It is known that heat (Q) required to change the phase of a given sample of mass m will be as follows.
........ (1)
where, = latent heat of fusion
Similarly, = mL_{v}[/tex] ....... (2)
where, = latent heat of vaporization
As it is known that value of latent heat of fusion of water is 334 J/g and latent heat of vaporization of water is 2260 J/g.
Hence, equating both the equations as follows.
=
=
=
= 54.13 g
Hence, we can conclude that 54.13 grams of ice will melt.
Answer:
0.4762 J/g°C.
Explanation:
<em>The amount of heat released to water = Q = m.c.ΔT.</em>
where, m is the mass of water (m = 15.0 g).
c is the specific heat capacity of water = ??? J/g°C.
ΔT is the temperature difference = (final T - initial T = 37.0°C - 30.0°C = 7.0°C).
<em>∴ The specific heat capacity of water = c = Q/m.ΔT</em> = (50.0 J)/(15.0 g)(7.0°C) = <em>0.4762 J/g°C.</em>
Answer:
B
Explanation:
Statement A
Vrms (root-mean-square velocity) is defined as follows.
Vrms = √(2RT/Mm), where R is the ideal gas constant, T is the temperature, and Mm is the molar mass of the gas. R and T are the same for both gases, but the molar mass is different, so Vrms for the two gases is different
Statement B
Kinetic energy KE of a molecule is related to temperature as follows:
KE = 3/2KbT, where Kb is the Boltzmann's constant and T is the temperature. The temperature is the same for both gases, so KE is the same for both.
Statement C
The rate depends on the number of molecules present. Since there are twice as many molecules of methane as propane, the rate is different for both gases.
Statement D
Pressure is directly proportional to the number of moles, based on the ideal gas law (pV = nRT). The volume and temperature are the same for both gases, so since there are twice as many molecules of methane as propane, the pressure is twice as great in the cylinder of methane.