Answer:
6.54% is the required concentration of lithium in an alloy (Al–Li).
Explanation:
Suppose 100 grams of an alloy of aluminium and lithium.
Density of an alloy = d 
Volume of an alloy = V

Volume of aluminum = 
Mass of aluminum = x
Density of aluminum = 

Volume of lithium = 
Mass of lithium = y
Density of lithium= 

..[1]


..[2]
Solving [1] and [2] :
we get :
x = 93.46 g
y = 6.54 g
Concentration of Li (in wt%) that is required:

That means that one mole of cobalt weighs 58.9332 grams (58.9332 g/mol). Based on that information, to convert 2 moles of cobalt to grams, we multiply 2 moles of cobalt by 58.9332.
The drug C3H30 is called allenolate / 1-hydroxycyclopropyl
weight is 55.06g/mol