Answer is: adding NaCl will lower the freezing point of a solution.
A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).
The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
Dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Answer:
pH of resulting solution = 7.98
Explanation:
The balanced equation
HA + NaOH - Na+ + A- + H2O
Number of moles of A = Number of moles of HA = Number of moles of NaOH
= 35.8/1000 * 0.020 = 0.000716 mol
Initial concentration of A = 0.000716/0.0608 = 0.01178 M
pKb = 14 – pKa = 14 -3.9 = 10.1
Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11
Kb = [HA][OH-]/[A-]
Kb = a^2/(0.01178 -a) = 7.943 * 10^-11
a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0
a = 9.673 * 10^-7
OH- = a = 9.673 * 10^-7 M
pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02
pH = 14-6.02 = 7.98
Answer:
1670 ml
Explanation:
molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity
Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.
