Answer:
The enthalpy of the reaction is coming out to be -380.16 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O%29%7D%29%2B%282%20mol%5Ctimes%5CDelta%20H_f_%7B%28H_2O%29%7D%20%29%5D-%5B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2H_4%29%7D%29%2B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O_4%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%2081.6%20kJ%2Fmol%29%2B2%20mol%5Ctimes%20-241.8%20kJ%2Fmol%29%5D-%5B%281%20mol%5Ctimes%20%2850.6%20kJ%2Fmol%29%29%2B%281%20mol%5Ctimes%20%289.16%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-380.16%20kJ)
Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.
1)
<span>m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = </span><span>58.5 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
</span>V(H2O) = 250ml = 250g = 0.25 kg<span>
</span><span>molality of NaCl:
</span><span>
n(NaCl)=m/M=1.95/58.5= 0.033 mole
</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation
Since NaCl dissociates into 2 ions in solution:
<span>
</span>milliosmolality of NaOH = 132 x 2 = 264 osmol<span>es/kg
</span>
2)
m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
V(H2O) = 250ml = 250g = 0.25 kg
molality of glucose:
n(gl)=m/M=9/180= 0.05 mole
molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg
milimolality of glucose = 0.132/0,001 = 200 mmole/kg
milliosmolality of glucose = milimolality x N of ions formed in dissociation
Since glucose does not dissociate, milimolality and milliosmolality are same:
milliosmolality of glucose = 200 osmoles/kg
3)
The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.
In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.
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