A) \[90.0\,g\,conc.HN{{O}_{3}}\]
B) \[70.0\,g\,conc.HN{{O}_{3}}\]
C) \[54.0\,g\,conc.HN{{O}_{3}}\]
D) \[45.0\,g\,conc.HN{{O}_{3}}\]
Correct Answer: D
Solution :
1000 mL has 2 moles \[250\,mL\,\,has\,\,\frac{2}{1000}\times 250=0.5\,\,moles\] \[\Rightarrow \,\,\,n=\frac{W}{M}\,\,\,W=n\times M=0.5\,\times 63\,\,g\] As 70 g of \[HN{{O}_{3}}\] is in 100 g solution So, \[0.5\,\,\times \,\,63\,g\,\,of\,\,HN{{O}_{3}}\] is in \[\frac{100}{70}\times 0.5\times 63\,\,=\,\,45\,g\] [We have assumed \[70%\] (mass by mass), although it should have been reported.]
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