# What mass of sodium chloride is required in the following scenario?

## A $350.00 \cdot L$ volume of $N a C l \left(a q\right)$ is required at $4.315 \cdot m o l \cdot {L}^{-} 1$ concentration. What mass of solute is required?

Apr 6, 2017

You would need over $88 \cdot k g$ of salt........

#### Explanation:

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

And thus $\text{moles of solute"="concentration"xx"volume of solution}$

And the problem gives us 2 of these 3 quantities, we just have to multiply and divide.

$\text{moles of solute} = 350.00 \cdot \cancel{L} \times 4.315 \cdot m o l \cdot \cancel{{L}^{-} 1} = 1510.25 \cdot m o l$.

And we then need to multiply this molar quantity by the molar mass of $N a C l$ in $g \cdot m o {l}^{-} 1$:

$1510.25 \cdot \cancel{m o l} \times \left(35.45 + 22.99\right) \cdot g \cdot \cancel{m o {l}^{-} 1} =$

??*g of salt..........