<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>
PH scale is from 1 to 14 and indicates how acidic or basic a solution is. To find pH or pOH we need to know the H⁺ ion concentration or OH⁻ concentration.
pH can be calculated using the following equation;
pH = -log[H⁺]
the H⁺ concentration of the given acid is 1.0 x 10⁻⁴ M. substituting this we can find the pH
pH = -log[1x10⁻⁴]
pH = 4
answer is 1) 4
