# What mass of sodium iodide is required to prepare a 250.0*mL volume of 1.50*mol*L^-1 concentration with respect to the salt?

Apr 8, 2017

Approx. $50 \cdot g$..............

#### Explanation:

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

Here $\text{concentration}$ $=$ $1.5 \cdot m o l \cdot {L}^{-} 1$, and $\text{volume of solution} = 250.0 \cdot m L$ or $250.0 \times {10}^{-} 3 L$.

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

$= 1.5 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 0.250 \cdot \cancel{L} = 0.375 \cdot m o l$ of $\text{NaI}$.

And mass of $\text{NaI}$ required $=$ $0.375 \cdot m o l \times 149.89 \cdot g \cdot m o {l}^{-} 1$

??*g