Johnny had a 50.0 mL sample of 0.4360 M NH 4 NO 3 that he diluted with water to
ID: 931194 • Letter: J
Question
Johnny had a 50.0 mL sample of 0.4360 M NH4NO3 that he diluted with water to a total volume of 250.0 mL. He has to figure out what is the ammonium nitrate concentration in the resulting solution? Go to this link https://youtu.be/MG86IFZi_XM
a) what equation is used? And what 2 pieces of glassware are used – explain what for
Johnny has to make the 0.1510 M NaOH solution needed for his lab. If he uses a 100.00mL volumetric flask, how many grams of NaOH does he need
b) explain how you would make this solution (so you will need to know grams of NaOH)
Explanation / Answer
According to law of dilution MV = M'V'
Where M = Molarity of stock = 0.4360 M
V = Volume of the stock = 50.0 mL
M' = Molarity of dilute solution = ?
V' = Volume of the dilute solution = 250.0 mL
Plug the values we get , M' = MV /V'
= ( 0.4360x50.0) / 250.0
= 0.0872 M
Therefore the the molarity of resulting ammonium nitrate solution is 0.0872 M
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Given Molarity of NaOH is , M = 0.1510 M
Volume of the solution , V = 100.00 mL = 0.100L
We know that Molarity , M = number of moles / volume in L
So number of moles of NaOH , n = Molarity x volume in L
= 0.1510 M x 0.100 L
= 0.0151 mol
Molar mass of NaOH = At.mass of Na + At.mass of O + At.mass of H
= 23+16+1
= 40 g/mol
We have number of moles , n = mass/molar mass
So mass of NaOH required , m = number of molesx molar mass
= 0.0151 mol x 40 (g/mol)
= 0.604 g
Exactly weigh & transfer 0.604 g of NaOH into a 100.0 mL volumetric flask & little amount of water , dissolve the contents & after complete dissolution add water upto the mark of the volumetric flask we get the required solution.
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