Answer:
The vapor pressure of the solution is 23.636 torr
Explanation:
Where;
is the vapor pressure of the solution
is the mole fraction of the solvent
is the vapor pressure of the pure solvent
Thus,
15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl
0.67 kg of water = [(0.67*1000)/(18)]moles = 37.222 moles of H₂O
Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)
Mole fraction of solvent (water) = (37.222)/(37.222+0.261)
Mole fraction of solvent (water) = 0.993
<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm
Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm
the vapor pressure of the solution = 0.0311 atm = 23.636 torr
Below are the steps to get the answers:
<span>1.) write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
2.) You are given everything needed to calculate
q=heat transfer=2.2*10^2, H3PO4 moles= 1.5*10^-3, NaOH moles=5.0*10^-3
3.) equation is deltaHneutraliztion=q/Moles of limiting reagent
H3PO4 is limiting reagent because lowest moles, and is used up first
4.) Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj, </span>
Initial Conditions:
Volume= v1= 417 cm³
Temperature= T1 = 278 K
Final Conditions:
Temperature= T2 = 231K
Volume = v2 =?
Use the general gas equation;
P1*v1/T1 = P2*v2/T2
As, the temperature is constant;
So,
v1/T1 = v2/T2
417/278 = v2/231
v2= 346.5 cm³
