The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
Answer: 
for the reaction is -186.75 J/K
Explanation:
Change in entropy () for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So = ![[3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol]](https://tex.z-dn.net/?f=%5B3%5Ctimes%2031.8%20J%2FK.mol%5D%2B%5B2%5Ctimes%20188.825%20J%2FK.mol%5D-%5B2%5Ctimes%20205.79%20J%2FK.mol%5D-%5B1%5Ctimes%20248.22%20J%2FK.mol%5D)
= -186.75 J/K
In the given situation, the gas is heated under constant volume. As energy is supplied to the system in the form of heat, the frequency of collision between the gas particles increases. This increases the temperature of the gas consequently bringing about a decrease in pressure.
Based on the ideal gas law:
PV = nRT
Here, P/T = nR/V
If P1, T1 and P2, T2 are the pressure and temperature values before and after heating respectively, then since nR/V is a constant in this case, we have
P1/T1 = P2/T2 which is the Gay-Lussac's law.
