# Awesome. Also need help on these two?

## Apr 2, 2018

So you want to make a $0.3 \cdot m o l \cdot {L}^{-} 1$ solution of $N a O H$ using $9.5 \cdot g$ of solute? I make it approx. an $800 \cdot m L$ volume.

#### Explanation:

Now, by definition, $\text{concentration"="moles of solute"/"volume of solution}$.

Now you might think I am being a bit anally retentive by including the units...so in fact I am, but this helps you to self-test the accuracy of your arithmetic...if the calculation gives appropriate units, the method of calculation was probably kosher. It is all too easy to divide when you should have multiplied, and vice versa.

And so filling the units in...

$0.30 \cdot m o l \cdot {L}^{-} 1 = \frac{\frac{9.5 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1}}{\text{Volume of solution}}$

And on rearrangement...

"Volume of solution"=((9.5*g)/(40.0*g*mol^-1))/(0.30*mol*L^-1)=??L

And for $b .$ we use the quotient again...

"Concentration"=((8.0*g)/(58.44*g*mol^-1))/(150*mLxx10^-3*L*mL^-1)=??*mol*L^-1...approx $1 \cdot m o l \cdot {L}^{-} 1$. Capisce?

Apr 2, 2018

a)
Given, $\text{Molarity} = 0.3 M$
and $\text{ given mass of } N a O H = 9.5 g$

$\text{Molarity" ="given mass"/ "molar mass " xx 1000/ "volume in mL}$

$0.3 = \frac{9.5}{40} \times \frac{1000}{\text{volume in mL}}$

$\text{volume in mL} = \frac{9.5}{40} \times \frac{1000}{0.3}$

$\text{volume in mL" =792" mL}$

b)

$\text{Molarity" ="given mass"/ "molar mass " xx 1000/ "volume in mL}$

$\text{Molarity} = \frac{8}{58.5} \times \frac{1000}{150} = 0.9 M$