# Given 1.0*mol*L^-1 solutions of NaCl(aq), what are the masses of solute in.....?

## $\left(i\right)$ $\text{a 1 L volume of solution;}$ $\left(i i\right)$ $\text{a 250 mL volume of solution;}$ $\left(i i i\right)$ $\text{a 100 mL volume of solution?}$

Sep 18, 2017

Well the molar mass of $N a C l$ is $\left(22.99 + 35.45\right) \cdot g \cdot m o {l}^{-} 1 =$ $58.44 \cdot g \cdot m o {l}^{-} 1$.......

#### Explanation:

From where did I get the atomic masses?

For the remaining problems, we use the relationship.....

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

And so, if we gots a $1 \cdot L$ volume of $1 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$ solution with respect there is a mass of.....

$1 \cdot L \times 1 \cdot m o l \cdot {L}^{-} 1 \times 58 , 44 \cdot g \cdot m o {l}^{-} 1 = 58.44 \cdot g$

And note that $1 \cdot {\mathrm{dm}}^{3} \equiv {\left(1 \times {10}^{-} 1 \cdot m\right)}^{3} \equiv 1 \times {10}^{-} 3 \cdot {m}^{3} = 1 \cdot L$, because there are $1000 \cdot L$ in a ${m}^{3}$.

And we also gots $250 \cdot c {m}^{3} \times {10}^{-} 3 \cdot {\mathrm{dm}}^{3} \cdot c {m}^{-} 3 \cdot 2 \cdot m o l \cdot {\mathrm{dm}}^{-} 3 \times 58.44 \cdot g \cdot m o {l}^{-} 1 = 29.22 \cdot g$

And I will let you the third.......