# What mass of solute is present in a 0.250*L volume of NaOH solution, for which [NaOH]-=2.50*mol*L^-1?

May 30, 2017

Approx. $25 \cdot g \ldots \ldots \ldots \ldots \ldots . .$

#### Explanation:

First we calculate the molar quantity present in the given volume of $N a O H$........

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

And thus $\text{Moles of solute"="Volume"xx"molarity}$.....

$250 \cdot \times {10}^{-} 3 \cdot \cancel{L} \times 2.5 \cdot m o l \cdot \cancel{{L}^{-} 1} = 0.625 \cdot m o l$, with respect to $N a O H$...............

which molar quantity corresponds to a MASS of..........

0.625*molxx40.0*g*mol^-1=??*g

You must simply KNOW that $1 \cdot m L \equiv 1 \times {10}^{-} 3 \cdot L$, i.e. the $\text{milli}$ prefix $\equiv {10}^{-} 3$.