Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L
Yes it is a exothermic reaction.
Huh I don’t get it what is this
Answer:
The molar mas of the X is 203.06 g/mol.
Explanation:
The pressure of the pure solvent = p = 23.8 Torr
Vapor pressure of the solution = ![p_s=23.5 Torr](https://tex.z-dn.net/?f=p_s%3D23.5%20Torr)
Mass of solute =14.4 g
Molar mass of solute = M
Moles of solute = ![n_1=\frac{14.4 g}{M}](https://tex.z-dn.net/?f=n_1%3D%5Cfrac%7B14.4%20g%7D%7BM%7D)
Mass of solvent or water = 100.0 g
Moles of water = ![n_2=\frac{100.0 g}{18 g/mol}=5.555 mol](https://tex.z-dn.net/?f=n_2%3D%5Cfrac%7B100.0%20g%7D%7B18%20g%2Fmol%7D%3D5.555%20mol)
Mole fraction of solute = ![\chi_1=\frac{n_1}{n_1+n_2}](https://tex.z-dn.net/?f=%5Cchi_1%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%7D)
The relative lowering in vapor pressure of the solution with non volatile solute is equal mole fraction of solute in solution.
![\frac{p-p_s}{p}=\chi_2=\frac{n_1}{n_1+n_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bp-p_s%7D%7Bp%7D%3D%5Cchi_2%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%7D)
![\frac{23.8 Torr-23.5Torr}{23.8Torr}=\frac{n_1}{n_1+5.555 mol}](https://tex.z-dn.net/?f=%5Cfrac%7B23.8%20Torr-23.5Torr%7D%7B23.8Torr%7D%3D%5Cfrac%7Bn_1%7D%7Bn_1%2B5.555%20mol%7D)
Solving for ![n_1](https://tex.z-dn.net/?f=n_1)
![n_1=0.070915 mol](https://tex.z-dn.net/?f=n_1%3D0.070915%20mol)
![n_1=\frac{14.4 g}{M}](https://tex.z-dn.net/?f=n_1%3D%5Cfrac%7B14.4%20g%7D%7BM%7D)
![M=\frac{14.4 g}{0.070915 mol}=20306 g/mol](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B14.4%20g%7D%7B0.070915%20mol%7D%3D20306%20g%2Fmol)
The molar mas of the X is 203.06 g/mol.