A $1.5 \cdot d m^{3}$ $1.5dm^3$ volume of $0.30 \cdot m o l \cdot d m^{- 3}$ $0.30moldm^-3$ $N a C l (a q)$ $NaCl(aq)$ was mixed with a $2.5 \cdot d m^{3}$ $2.5dm^3$ volume of $0.70 \cdot m o l \cdot d m^{- 3}$ $0.70moldm^-3$ $N a C l (a q)$ $NaCl(aq)$ . What is the final concentration of $N a C l (a q)$ $NaCl(aq)$ ?

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A $1.5 \cdot d m^{3}$ $1.5dm^3$ volume of $0.30 \cdot m o l \cdot d m^{- 3}$ $0.30moldm^-3$ $N a C l (a q)$ $NaCl(aq)$ was mixed with a $2.5 \cdot d m^{3}$ $2.5dm^3$ volume of $0.70 \cdot m o l \cdot d m^{- 3}$ $0.70moldm^-3$ $N a C l (a q)$ $NaCl(aq)$ . What is the final concentration of $N a C l (a q)$ $NaCl(aq)$ ?

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Answer 1 · 2016-10-27T20:38:39

$[N a C l (a q)] = 0.55 \cdot m o l \cdot L^{- 1}$ $[NaCl(aq)]=0.55*mol*L^-1$ .

Explanation:

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

We have to calculate the quantity of all the ions in solution, and then divide this by the total volume.

$Moles of NaCl(i)$ $"Moles of NaCl(i)"$ $=$ $=$ $1.50 \cdot d m^{3} \times 0.30 \cdot m o l \cdot d m^{- 3} = 0.45 \cdot m o l$ $1.50*dm^3xx0.30*mol*dm^-3=0.45*mol$

$Moles of NaCl(ii)$ $"Moles of NaCl(ii)"$ $=$ $=$ $2.50 \cdot d m^{3} \times 0.70 \cdot m o l \cdot d m^{- 3} = 1.75 \cdot m o l$ $2.50*dm^3xx0.70*mol*dm^-3=1.75*mol$

And thus $[N a C l]$ $[NaCl]$ $=$ $=$ $\frac{0.45 \cdot m o l + 1.75 \cdot m o l}{4.0 \cdot d m^{3}}$ $(0.45*mol+1.75*mol)/(4.0*dm^3)$

$=$ $=$ $0.55 \cdot m o l \cdot d m^{- 3}$ $0.55*mol*dm^-3$ .

Note that $1 \cdot d m^{3}$ $1*dm^3$ $=$ $=$ ${(10^{- 1} m)}^{3}$ $(10^-1m)^3$ $=$ $=$ $10^{- 3} m^{3} = 1 L$ $10^-3m^3=1L$ . We have (reasonably) assumed the volumes of solution to be additive.

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