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?

2 Answers
Apr 17, 2017

.143 moles

Explanation:

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.

Apr 17, 2017

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

Explanation:

"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?