What mass of glucose, #C_6H_12O_6#, would be required to prepare #5 * 10^3# #L# of a #0.215# #M# solution?

1 Answer
Feb 29, 2016

Answer:

193661.25g #C_6H_12O_6# (before sig figs)

Explanation:

Molarity is defined as #M#= #(mols)/(1L solution)#

from this, we can deduct that .215#M# #C_6H_12O_6#= #(.215mol)/(1L solution)#

#C# = 12.01 * 6 = 72.06g
#H# = 1.0079 * 12 = 12.0948g
#O# = 16.00 * 6 = 96.00g
#C_6H_12O_6#= 180.15g

Start with what is given

#5*10^3##L# #C_6H_12O_6# #(.215 molC_6H_12O_6)/(1L solution)# #(180.15g)/(1mol C_6H_12O_6)#= 193661.25g #C_6H_12O_6# (before sig figs)