Post a New Question

chemistry

posted by .

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

  • chemistry -

    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.

  • chemistry -

    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
    a)100g
    b)100 mg
    c) 5.85 g
    d) 58.5g

    the answer is c

  • chemistry -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. science

    Q4.you will be asked to prepare 25.0 mL of a 0.019 M solution of sodium chloride during this experiment.... a.calculate mass of sodium chloride needed to prepare solution. (0.25L)(0.019M NaCl/L)*(58.5g NaCl/1mole NaCl)=0.278 g NaCl....is …
  2. chemistry

    The traditional method of analyzing the amount of chloride ion present in a sample was to dissolve the sample in water and then slowly add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight …
  3. Chemistry

    A buffer solution of pH=9.24 can be prepared by dissolving ammonia and ammonium chloride in water. How many moles of ammonium chloride must be added to 1.0 L of .50 M ammonia to prepare the buffer?
  4. Chemistry Balancing Equations

    Can you please tell me if these are correct?
  5. Chemistry

    How will the boiling point of a liter of water containing 1 mole of sodium chloride (NaCl) compare with that of a liter of water containing 1 mole of calcium chloride (CaCl2)?
  6. chemistry

    30. How will the boiling point of a liter of water containing 1 mole of sodium chloride (NaCl) compare with that of a liter of water containing 1 mole of calcium chloride (CaCl2)?
  7. Chemistry

    1) Calculate the amount of sodium chloride, NaCL (mol wt 58.44) and potassium chloride; KCL (Mol wt 74.56) needed to prepare 500 ml of 1.25 NaCL/10 mM kcl solution dilutions in water.
  8. chemistry

    a solution (S) is prepared by dissolving 50.0 g calcium chloride in 250 ml volumetric flask wich is then filled with distilled water up to the line mark. calculate the molar concentration of solution (S)
  9. Chemistry

    a solution is prepared by dissolving 50 g of calcium chloride in a 250ml volumetric flask which is then filled with distilled water up to the line-mark. Calculate the molar concentration of the solution.
  10. Chemistry

    Help please. Am doing a past exam paper and I hit a road block with this question: Chloride in a brine solution is determined by a precipitation titration. A 10.00 mL aliquot of the solution is titrated with 15.00mL of standard 0.1182 …

More Similar Questions

Post a New Question