A) \[{{P}_{4}}{{S}_{3}}\]
B) \[{{P}_{2}}{{S}_{2}}\]
C) \[P{{S}_{2}}\]
D) \[{{P}_{2}}{{S}_{4}}\]
Correct Answer: A
Solution :
Choice A is \[{{P}_{4}}{{S}_{3}}\] \[\because \] \[\frac{31\times 4}{(124)}gm\]P is present in 220gm \[{{P}_{4}}{{S}_{3}}\] \[\therefore \] 1.24gm P is present in = \[\frac{220}{124}\times 1.24=2.2gm\]
You need to login to perform this action.
You will be redirected in
3 sec
