Density = mass/volume = 2000/4000 = 0.5 grams/cm3. Hope this hopes!
The standard Gibbs free energy of formation ΔGf° of Rb(s), H2(g) and Pb(s) are all zero. Similar to enthalpies of formation, the values of the standard Gibbs energies of formation are zero for the elements in their most stable forms at room conditions 298 Kelvin and one atmosphere pressure.
Answer : q = 6020 J, w = -6020 J, Δe = 0
Solution : Given,
Molar heat of fusion of ice = 6020 J/mole
Number of moles = 1 mole
Pressure = 1 atm
Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.
The relation between heat and molar heat of fusion is,
(in terms of mass)
or,
(in terms of moles)
Now we have to calculate the value of q.
![q=6020J/mole\times 1Mole=6020J](https://tex.z-dn.net/?f=q%3D6020J%2Fmole%5Ctimes%201Mole%3D6020J)
When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.
So, the value of ![\Delta e=0](https://tex.z-dn.net/?f=%5CDelta%20e%3D0)
Now we have to calculate the value of w.
Formula used : ![\Delta e=q+w](https://tex.z-dn.net/?f=%5CDelta%20e%3Dq%2Bw)
where, q is heat required, w is work done and
is internal energy.
Now put all the given values in above formula, we get
![0=6020J+w](https://tex.z-dn.net/?f=0%3D6020J%2Bw)
w = -6020 J
Therefore, q = 6020 J, w = -6020 J, Δe = 0
Freezing point depression is the phenomena that describes why adding a solute to a solvent results in the lowering of the freezing point of the solvent. ... Adding solute to a solvent will essentially dilute the solvent molecules, and according to Raoult's law, this leads to a decrease in vapor pressure.