If the percentage of nitrogen in an organic compound is 125%, then how much of the organic compound should be taken so as to produce 50 mL of `N_(2)`
If the percentage of nitrogen in an organic compound is 125%, then how much of the organic compound should be taken so as to produce 50 mL of `N_(2)` at 300 K and 715 mm pressure (Aq. tension = 15 mm).
A. 0.419 g
B. 0.149 g
C. 0.914. g
D. 0.941 g
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Correct Answer - A
`V_(1) = 50` mL `" " V_(2) = ?`
`P_(1) = 715 - 15 = 700` mm `P_(2) = 760` mm
`T_(1) = 300 K " " T_(2) = 273 K`
`V_(2) = (P_(1)V_(1)T_(2))/(P_(2)T_(1)) = (700 xx 50 xx 273)/(760 xx 300) = 41*9` mL
22400 mL of `N_(2)` at NTP has mass = 28 g
`41*9` mL of `N_(2)` at NTP has mass
= `(28)/(22400) xx 41*9 = 0*05` g
`12*5` g of nitrogen in present in O.C. = 100g
`0*05` g of nitrogen is present in O.C.
= `(100 xx 0*05)/(12*5) = 0*419` g
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