A sample of 6.40 g of NaCl was dissolved in water and made up to the volume in a

Question

A sample of 6.40 g of NaCl was dissolved in water and made up to the volume in a 250-mL volumetric flask. A 10.00 mL sample of this solution was transferred to another 250-mL. volumetric flask and diluted to the mark with water.(a) Determine the concentration (in M) of the final solution. (b) If instead of doing a serial dilution as described in question "a" you want to make the solution directly from solid NaCl, calculate the mass of NaCl needed to make 100.00 ml, of the final solution. (c) In 20 words or less, describe how you would prepare the solution in question "b" using solid NaCl.

Explanation / Answer

weight of the sample taken = 6.40 gm

molecular weight of NaCl = 58.5 gm

mole of NaCl = weight taken / molecular weight = 6.40 gm / 58.5gm = 0.109 mole

molarity (M) = mole / solution in L

here volume of solution = 250 ml = 0.25 L

therefore, mole = 0.109 / 0.25 = 0.436 M

(a) initial volume (V1) = 10 ml

initial strength (s1) = 0.436 ml

final volume (V2) = 250 ml

final strength (S2) = ?

stabdard relation for dilution...

S1 * V1 = S2 * V2

or, S2 = S1*V1 / V2

or, S2 = (10*0.436) / 250 = 0.017 M

(b) preparation of 100 ml 0.017 M NaCl solution.

as per standard relation..

molarity (M) = mole / solution in L

mole of NaCl = molarity * solution in L

or, mole of NaCl = 0.017 mole / L * 0.1 L = 0.0017 mole

weight of NaCl = 0.0017 mole * molecular weight of NaCl ( 58.5 gm) = 0.099 gm

(c) Take 0.099 gm NaCl in a 100 ml volumetric flask and then water is added upto the mark.

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