posted by james Neil on .
a chloride standard containing 100mmol/L of chloride ion can be prepared by dissolving the following amount of NACL in 1000ml of distilled water.
dont know how to prepare a standard
but heres my math
I don't see any math. NACL and NaCl are not the same thing. And placing it IN 1000 mL of water is not the way to go if you want a standard of 150 mmoles/L. xxmoles/L are prepared by placing the solid in a volumetric flask, adding a little water, dissolving the solid completely, then making the final volume to the mark in the volumetric flask. That way you have 1000 mL of SOLUTION; the other way you have more than 1000 mL of solution.
sorry, don't undestand how to make a standard. This is the problem
heres the math
100mmol x 1mol/1000mmol x 58.44g/m= 5.85g of NACL
throw 5.58 gram of nacl into a thousand water? or is it 994.42 water ?>
i have 4 choices for answers
c) 5.85 g
the answer is c
Technically, none of the answers listed are correct. I expect the intent of the problem is to answer with 5.85. But here is how you do it (and you transposed the numbers in your math but otherwise you math is good).
100 mmoles x (1 mole/1000 mmols) x (58.44 g/m) = 5.84 grams NaCl.
Here is how you make up the solution. Carefully weigh out 5.84 g dried NaCl, place into a 1000 mL volumetric flask, add some water, swirl until all of the solid is dissolved, then add distilled water to make to the mark in the volumetric flask. That way you will have 5.84 g NaCl/1000 mL of solution and that will be 100 mmols/L of solution. I suspect the author of the problem rounded for the molar mass of NaCl as 58.5 and perhaps may not realize that adding 5.85 g NaCl to 1000 mL water will produce a final volume of a little over 1000 mL; therefore, the concn will be a little less than 100 mmoles/L.