3% Ni in steel to be analyzed from 1g of
steel sample. What volume of 1% wt of DMG
(dimethyl glyoxime) in alcohol to be used
to provide a 50% excess of DMG for the
analysis. The density of alcohol (0.79 g/ml).
FM : Ni(58.69); DMG(116.12).
This is the problem.
What I have done is:
1st I've clculated that .03 g Ni in present in 1g steel then from the reaction between Ni-DMG I calculated that 0.1187 g DMG is required for .03g Ni (as 2 mol of DMG is needed for 1mol of Ni). Then adding 50% extra it becomes .1780g. Now 1g DMG is present in 100g solution, that means 99g alcohol, from here vol of 99g alcohol can be calculated using its density. Now 1g DMG is present in that much alcohol so .178 g is present in how much. Thus I got the result 22.25ml
Now I want to confirm it. Please help me by telling whether the procedure is right or wrong and if it is wrong in any step please help me to rectify it by explanig.