How many moles of sodium hydroxide is contained in a #65.0mL# volume of #"sodium hydroxide"# that is #2.20mol*L^-1# with respect to the base?

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Answer 1 · 2017-04-17T14:05:16

.143 moles

We're going to use this equation

#n=M*V#

M meaning our molarity, n meaning our moles involved, and v meaning our volume in Liters.

#n=2.20 * .065#

n=.143 moles.

Answer 2 · 2017-04-17T14:08:19

Approx. #0.14*mol...............#

#"Concentration"="Moles of solute"/"Volume of solution"#.

And thus #"Moles of solute"="Concentration"xx"Volume of solution"#

And so we take the product..........

#2.20*mol*cancel(L^-1)xx65.0*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)=0.143*mol# with respect to #NaOH#.

Why is the answer consistent dimensionally?

How many moles of sodium hydroxide is contained in a #65.0*mL# volume of #"sodium hydroxide"# that is #2.20*mol*L^-1# with respect to the base?