What mass of barium chloride is required to make a 100*cm^3 volume of 0.25*mol*L^-1 solution with respect to BaCl_2(aq)?

Jun 24, 2017

Approx. $5 \cdot g$.........

Explanation:

$\text{Concentration"="Moles of solute (mol)"/"Volume of solution (L)}$

And thus the product $\text{volume"xx"concentration"="moles of solute.}$

And thus..........

${n}_{B a C {l}_{2}} = 0.25 \cdot m o l \cdot {L}^{-} 1 \times 100 \cdot c {m}^{3} \times {10}^{-} 3 \cdot L \cdot c {m}^{-} 3$

$= 0.025 \cdot m o l$ $B a C {l}_{2}$

And since $\text{no. of moles"="mass"/"molar mass}$

"mass"=0.025*cancel(mol)xx208.23*g*cancel(mol^-1)=??g

All I have done here is to use the given quotient that defines $\text{concentration}$, and divided or multiplied to give the quantity I want. As a check on my arithmetic, I included the units of each quantity. The fact that I clearly got an answer in $\text{grams}$ for a question that asked for an answer in $\text{grams}$ persuades me that I got the order of operations right (for once!).