Answer:
π = 14.824 atm
Explanation:
wt % = ( w NaCL / w sea water ) * 100 = 3.5 %
assuming w sea water = 100 g = 0.1 Kg
⇒ w NaCl = 3.5 g
osmotic pressure ( π ):
∴ T = 20 °C + 273 = 293 K
∴ C ≡ mol/L
∴ density sea water = 1.03 Kg/L....from literature
⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln
⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol
⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M
⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K
⇒ π = 14.824 atm
The empirical formula for the unknown compound would be: C2H4O (2 molecules of Carbon, 4 molecules of Hydrogen, and 1 molecule of Oxygen)
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
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^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
Hence, the correct answer is Option D.