[browniee]

This is a question from one of my practice papers.
6.853g of sodium cyanide was dissolved in water and brought to a volume in a 100ml volumetric flask. The final mass of the solution was found to be 107.166g.

Calculate the composition of the solution expressed as :
1) Moles, 2) Molarity, 3) Normality, 4) weight/volume %, 5) weight/weight %

For (1) -> Mole composition = moles of NaCN / moles of total mixture
Moles of 6.853g of NaCN = 6.853/49.008 = 0.13983 moles
Mass of remaining water = 107.166 - 6.853 = 100.313g
Moles of this water = 100.313/18.0148= 5.56837 moles
Total moles = 0.13983 + 5.56837 = 5.7082 moles
mole composition = 0.13983/5.7082 = 0.0245

For (2) -> using the same method, i tried to do it but struggled.
Molarity composition = molarity of NaCN / Molarity of solution

Molarity of 6.853g of NaCN that went to the water = 0.13983 /  0.1 = 1.3983 M
Then i got stucked here, arent molarity of 6.853g of NaCn that went into the water same as molarity of the total solution ? Nevertheless i still did it.
Molarity of total solution = 5.7082 / 0.1 = 57.082M
Molarity composition = 1.3983/57.082

For (3) ->
i only know how to calculate the normality of 6.853g of NaCN -> 1.3983 M = 1.3983 x 1 = 1.3983 N
but i got stuck at the normality of total solution

For (4) -> weight/volume % = mass of solute / volume of total solution x 100% .
i just substituted in the relevant numbers.

For (5) -> weight/weight % = mass of solute / mass of total solution x 100%

i believe i am wrong some where. especially from (1)-(3) as when i did it, i did not think it was right. am i correct on my concept? or which part should i make changes? Thanks !

[Borek]

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