Freezing point depression depends of the number of particles of the solute in the solution.
1)Pure water have highest freezing point. All other solutions with given solutes will have lower temperatures.
2) The more particles of the solute in the solution the lower freezing point is going to be.
<span>b. 1.0 m NaCl ( dissociates and give 2 mol ions (1 mol Na⁺ and 1 mol Cl⁻))
c. 1.0 m K3PO4 (</span>dissociates and give 4 mol ions (3 mol K⁺ and 1 mol PO4³⁻)<span>
d. 1.0 m CaCl2 (</span>dissociates and give 3 mol ions (1 mol Ca²⁺ and 2 mol Cl⁻))<span>
e. 1.0 m glucose (c6h12o6) (glucose does not dissociate, and solution have
1 mole of particles of the solute(glucose))
The largest number of particles has </span>1.0 m K3PO4 solution, and it is has lowest freezing point . Answer is C.
Answer:
The answer to your question is: T2 = 235.44 °K
Explanation:
Data
V1 = 3.15 L V2 = 2.78 L
P1 = 2.40 atm P2 = 1.97 atm
T1 = 325°K T2 = ?
Formula
Process
T2 = (P2V2T1) / (P1V1)
T2 = (1.97x 2.78x 325) / (2.40 x 3.15)
T2 = 1779.895 / 7.56
T2 = 235.44 °K
Its a compound because they come together
Answer:
0.40 g/cm3
Explanation:
density = mass / volume.
mass = 65.2 grams
volume = 10*1.1*15=165 cm3
so density = 65.2/165=0.40 g/cm3
