<span>1.16 moles/liter
The equation for freezing point depression in an ideal solution is
ΔTF = KF * b * i
where
ΔTF = depression in freezing point, defined as TF (pure) ⒠TF (solution). So in this case ΔTF = 2.15
KF = cryoscopic constant of the solvent (given as 1.86 âc/m)
b = molality of solute
i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1.
Solving for b, we get
ΔTF = KF * b * i
ΔTF/KF = b * i
ΔTF/(KF*i) = b
And substuting known values.
ΔTF/(KF*i) = b
2.15âc/(1.86âc/m * 1) = b
2.15/(1.86 1/m) = b
1.155913978 m = b
So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
The greatest aqueous freezing point is (D) 0.10 KCI
The volume of a gas that its pressure increase to 3.4 atm is calculated as follows
By use of boyles law that is P1V1=P2V2
V1=4.0 L
P1=1.1 atm
P2=3.4 atm
V2= P1V1/P2
(1.1 atm x 4.0 L)/3.4 atm= 1.29 L
A) 120 mm
B) 127 mm
C) 914.4 mm
D) 1000 mm
E) 3048 mm
