Pre-Lab Questions 1. Calculate the molarity (M) and molality (m) of a solution c

Question

Pre-Lab Questions 1. Calculate the molarity (M) and molality (m) of a solution consisting of 1.56 g of sodium chloride dissolved in 10.00 ml of acetone (density of acetone o.791 g/mL). Show all work. 2. Why is it important to specify units when calculating concentration? 3. The solvent used during this experiment is glacial acetic acid (CH3CooH). Look up the freezing point depression constant (Kf) for this solvent and the normal freezing point (in oC) of acetic acid make sure to cite where this information was obtained

Explanation / Answer

1 ) weight of NaCl= 1.56 g

   molecular weight NaCl = 58.5 g / mol

moles of NaCl = 1.56 / 58.5

                        = 0.027

volume of acetone = 10 ml

density of acetone = 0.791 g / ml

density = weight / volume

weight = density x volume

           = 0.791 x 10

           = 7.91 g

molality : the number of moles of solute present in 1 Kg of solvent is know as molality

molality (m) = (moles / solvent weight (g) ) x 1000

                   = ( 0.027 / 7.91 ) x 1000

                    = 3.41m

molality = 3.41 m

relation between molarity and molality :

m = 1000 M / (1000 d - M x molar mass of solute)

3.41 = 1000 M / (1000 x 0.791 - M x 58.5)

M = 2.25

molarity = 2.25 M

2)

unit are very important in case of concentration calculation. we missplace any unit there be will many problems arises.

suppose a patient needs to take 2 M antacid for reduce acidity in the stomach. insted of 2M if he take 2M antacid there will be proble again.

2M antacid may or may not equal to 2N . so the patient will face again problem.

3)

for galcial acetic acid freezing point depression constant (Kf) = 3.90 0C / m

acetic acid freezing point (Tf) = 16.60C

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