Describe how you would prepare 150.0 mL of a 0.250 M NaOH solution given that th
ID: 998507 • Letter: D
Question
Describe how you would prepare 150.0 mL of a 0.250 M NaOH solution given that the NaOH is supplied as a solid. b) In the laboratory a stock solution of glacial acetic acid has concentration of 17.4 M. A student was asked to prepare 5.000L of 0.500 M CH3COOH (acetic acid) from the stock solution. Would the student be correct in diluting 150.00 mL of the stock solution to a total volume of 5.000L? Explain your answer, providing correct volumes if necessary. c) What is the benefit of placing salt on roads in extremely cold weather? Explain your answer on a molecular basis. (Pls no more than a page long answer)Explanation / Answer
Exact molar mass of NaOH = 39.98 g/mol.
I.e. On dissolving 39.98 g of solid NaOH in some distilled water and then diluting to 1000 mL by adding distilled water the solution formed will be 1M.
Hence we write,
If, 1 M 1000 mL NaOH solution 39.98 g NaOH.
Then, 0.250 M 1000 mL NaOH solution say ‘A’ g NaOH.
A x 1 = 0.250 x 39.98
A = 9.995 g NaOH.
It means, for preparing 0.250 M NaOH
If, 1000 mL NaOH 9.995 g NaOH
Then, 150 mL NaOH ‘M’ g NaOH
M x 1000 = 9.995 x 150
M = (9.995 x 150)/1000
M = 1.449 g NaOH.
Procedure:
Dissolve 1.449 g NaOH in some amount of deionized distilled water and then dilute it in 150 mL volumetric flask by adding appropriate amount of distilled water.
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b)
Molar equation M1V1 = M2V2 can be used here.
Let us assume that,
For the Acetic acid solution to be prepared, M1 = 0.500M and V1 = 5.000 L
For the stock acetic acid solution, M1 = 17.4 M and V2 = ?
Let use above molar equation,
M2 x V2 = M1 x V1
17.4 x V2 = 0.500 x 5.000
17.4 x V2 = 2.500
V2 = 2.500 / 17.4
V2 = 0.1437 L
V2 = 0.14368 x 1000 mL
V2 = 143.68 mL
V2 = 143.68 mL
150.00 mL is the wrong volume of stock solution.
143.68 mL of 17.4 M stock solution will be required to prepare 5.000 mL 0.500 M AcOH solution.
143.68 mL of stock AcOH solution firstly added to some amount of AcOH and then this solution is finally diluted to 5.000 L.
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c)
Salt like NaCl is placed on road in extremely cold winters to avoid formation of ice which blocks the roads.
Here the colligative property of solution namely depression in freezing point phenomenon plays an important role.
Addition of NaCl a strong electrolyte to water depresses it’s freezing point and ice freezing prevented.
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