1.

1.525g of an organic compound was Kjeldahlised and the ammonia so produced was passed into 30mL of N HCl solution. The remaining HCl was further neutralised by 120mL of (N)/(10) NaOH solution. Calculate the percentage of nitrogen in the compound

Answer»

Solution :m.e. of `NaOH = (1)/(10) XX 120= 12`
m.e of remaining HCl= 12
m.e. of HCl (total) `=1 xx 30= 30`
m.e. of HCl neutralised by `NH_(3)= 30-12=18`
`therefore` m.e. of `NH_(3)= 18`
Eq of `NH_(3)= (18)/(1000)= 0.018`
`therefore` mole of `NH_(3)= 0.018`
Now mole of N in `NH_(3) = 1 xx` mole of `NH_(3)`
`=1 xx 0.018 = 0.018`
Weight of nitrogen (N) = `0.018 xx 14= 0.252g`
PERCENTAGE of nitrogen `= (0.252)/(1.525)xx 100= 16.52%`


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