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)

Answer 1

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.