Here we have a dilution chemistry problem. For problems like this one, we use the following:
#color(white)(aaaaaaaaaaaaaa)color(purple)(M_1V_1 = M_2V_2#
Where
#M_1 = "molarity of solution 1"# #(3.0" M "HNO_3)#
#V_1 = "volume of solution 1 (??)"#
#M_2 = "molarity of solution 2"# #(0.250" M "HNO_3)#
#V_2 = "volume of solution 2 (0.040 L)"#
We know that molarity is defined as moles of solute per liter of solution.
#color(white)(aaaaaaaaa)"Molarity" = "moles of solute"/(1" Liter solution")#
The question
#color(white)(----)#How would you prepare 40.0 mL of 0.250 M HNO3 from a stock (initial) solution of 3.00 M HNO3
is just asking you how would one make a diluted solution of #HNO_3# from a concentrated stock solution of #HNO_3#.
Steps
-
#M_1V_1 = M_2V_2#
-
#(3.0" M")(V_(1)) = (0.250" M")(0.040" L")#
-
#V_(1) = [(0.250cancel" M")(0.040" L")]/(3.0cancel" M")#
-
#V_(1) = 0.0033" L" or 3.3" mL"#
This means you would need #3" mL"# of the concentrated stock solution of #HNO_3# and then you would need to dilute it down with water until you reach the #40" mL"# mark.
