# 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?

Apr 17, 2017

.143 moles

#### Explanation:

We're going to use this equation

$n = M \cdot V$

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

$n = 2.20 \cdot .065$

n=.143 moles.

Apr 17, 2017

Approx. $0.14 \cdot m o l \ldots \ldots \ldots \ldots \ldots$

#### Explanation:

$\text{Concentration"="Moles of solute"/"Volume of solution}$.

And thus $\text{Moles of solute"="Concentration"xx"Volume of solution}$

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

$2.20 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 65.0 \cdot \cancel{m L} \times {10}^{-} 3 \cdot \cancel{L} \cdot \cancel{m {L}^{-} 1} = 0.143 \cdot m o l$ with respect to $N a O H$.

Why is the answer consistent dimensionally?