Answer: a. 0.75m (NH4)3PO4 will have the lowest freezing point .
Explanation: Freezing point decreases as the concentration of the solute substance increases. Assuming the same solvent for all of them, for instance water.
∆T= i.Kf.b
∆T= freezing point depression
i= vant Hoff factor
Kf= molality
Assuming water to be the solvent for all Kf=1.86°C/M
VANT HOFF FACTORS :
For (NH4)3PO4
This has 3 ionic bonding and 1 covalent bonding making it 4 bond
Therefore i=4
For CaSO4
This has 1 metallic bond and 1 covalent bond making it 2 bond.
Therefore i=2
For LiCl
This has 1 metallic bond and 1 non metallic bond making it 2 bond.
Therefore i=2
For CH3OH
This has only 1 covalent bond.
Therefore i=1
MOLALITY:
(NH4)3PO4 = 0.75M
CaSO4= 1.0M
LiCl= 1.0M
CH3OH= 1.5M
FREEZING POINT DEPRESSION:
For (NH4)3PO4
∆T= 4×0.75×1.86=5.58°C
For CaSO4
∆T= 2×1.0×1.86=3.72°C
For LiCl
∆T = 2×1.0×1.86= 3.72°C
For CH3OH
∆T= 1×1.5×1.86=2.79°C
REMEMBER THE HIGHER THE FREEZING POINT DEPRESSION THE LOWER THE FREEZING POINT.
FREEZING POINT DEPRESSION IS THE CHANGE IN THE FREEZING POINT PROPORTIONAL TO THE AMOUNT OF SOLUTE ADDED THE THE SOLUTION.
THEREFORE THE ONE WITH THE LOWEST FREEZING POINT IS (NH4)3PO4
Answer: Laws have been passed banning the production of a living copy of a person.
i just took the test so i know the answer is correct.
As given:
Initial moles of P taken = 2 mol
the products are R and Q
at equilibrium the moles of
R = x
total moles = 2 + x/2
Let us check for each reaction
A) P <-> 2Q+R
Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = 2x
moles of R = x
Total moles = (2-x) + 2x + x = 2 +2x
B) 2P <-> 2Q+R
Here x moles of P will give x moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x
moles of R = x/2
Total moles = (2-x) + x + x/2 = 2 + x/2
C) 2P <-> Q+R
Here x moles of P will give x/2 moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x /2
moles of R = x/2
Total moles = (2-x) + x + x = 2
D) 2P <-> Q+2R
Here x moles of P will give x/2 moles of Q and x moles of R
So at equilibrium
moles of P left = 2-x
moles of Q = x/2
moles of R = x
Total moles = (2-x) + x/2 + x = 2 + x/2
Answer:
K3PO4
Explanation:
Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;
SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)
C4H11N (not ionic in nature hence it can not dissociate into ions)
K3PO4-------> 3K^+ + PO4^3- (4 particles)
Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)
Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.
