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Submitted by shboulav on Sun, 09/11/2011 - 17:34

Explain the steps you would take to prepare 0.500 L of a 0.250-M solution of aluminum nitrate?

 

How many grams of copper (II) chloride are needed to prepare 100.0 mL of a 0.120-M solution?

 

What is the molality of a 0.250-M solution of sodium chloride if the density is 1.015 g/mL? Carry the units.

Steps to prepare 0.500 L of a 0.250M solution of aluminum nitrate

I'll pretend this is for a real lab, just in case.

Step 1 _ Look for aluminum nitrate on the chemicals shelves. You may find the anhydrous form or a hydrate. Depending on what you find, you would find out or calculate the molar mass.

Step 2 _ Find the molar mass. Reagent obtained from a commercial provider will most likely have a formula weight or molecular weight listed on the label. As this is probably a classroom problem, you have to find it yourself. Physics and chemistry handbooks list formula weights, and it's easy to find them online, but your instructor probably wants you to add atomic weights. It should be 213.0g/mol (anhydrous)
375.1 g/mol (nonahydrate)

Step 3 _ Find how many moles you need for 0.500L of solution. That would be (0.250mol/L)(0.500L).

Step 4 _ Calculate the mass in grams that corresponds to that many moles.

Step 5 - Weigh the amount calculated and transfer it into a 500 mL class A volumetric flask.

Step 6 _ Add water (maybe 300-400mL) and swirl to dissolve. If needed, I would add a magnetic stir bar and stir on a stirplate.

Step 7 _ Make up to volume (we call it q.s.) by adding enough water to get to the mark. Then stopper tightly and mix by inverting the flask repeatedly.

Step 8 _ Label it so as to satisfy all internal and external rules and regulations (probably description, date prepared, preparer, storage conditions, expiration date, lot numbers, notebook reference).

How many grams of copper (II) chloride ...You'll get the idea from the answer to the previous question.

What is the molality of a 0.250-M solution of sodium chloride if the density is 1.015 g/mL? Carry the units.

Mass of NaCl in 1L solution = (0.250mol/1000mL)(1000mL)(58.44g/1 mol) = (0.250mol)(58.44g/1 mol) = 14.61 g

Mass of 1L solution = (1000mL)(1.015g/mL) = 1015g

Mass of water in 1L solution = 1015g - 14.61g = 1000g = 1kg

Since you have 0.250mol NaCl dissolved in 1.000kg water, your molality is

m = 0.250mol/1.000kg = 0.250mol/kg = 0.250m

Submitted by KMST on Sun, 09/11/2011 - 19:50 Permalink