The Stock A sowing information pertains to Problems 1-3: solution is prepared by

Question

The Stock A sowing information pertains to Problems 1-3: solution is prepared by dissolving 3.213 g of ammonium chlorate in enough water to 00 mL of solution, A second solution is then prepared by diluting a 5.0 mLaliquot mple) of the stock solution to 250 mL·The density of the new solution Sa is 1.00 g/mL. What is the molar concentration of the solute in the new solution? mel Vi-. 50 mL_ ?.scoL 2. V2. = 12.7 2. How many grams of solute are in 20 ?L of the new solution? miero litor mass ? .co alot . O.02 mL 0.02 g

Explanation / Answer

3.213 g of NH4ClO4 in 500 mL solution,

Molar mass of NH4ClO4 = 117.4891 g/mol.

Number of moles = 3.213/117.4891 mole

= 0.027 mole

In 1000 mL solution, number of moles = 2*0.027

= 0.054 moles.

So, the molarity of the stock solution = 0.054 M

1)

5 ml of stock solution is diluted to 250 mL, so dilution factor = 250/5 = 50

So, molarity of the new solution = 0.054/50 M = 0.0011 M

2)

Molarity, of the solution = 0.0011 M,

so 20 ?L (micro liter) will contain =0.0011*20*10^(-6)*117.4891

= 2.584 ?g (micro gram)

3)

500 mL solution contains = 3.213 g of NH4ClO4

5 ml contains = 0.03213 g, diluted to 250 ml (d=1g/ml). So, 100 g contains

= 0.03213/2.5 = 0.013%

So, wt percentage = 0.013%

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