A $1.5dm^3$ volume of $0.30moldm^-3$ $NaCl(aq)$ was mixed with a $2.5dm^3$ volume of $0.70moldm^-3$ $NaCl(aq)$ . What is the final concentration of $NaCl(aq)$ ?

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Answer 1 · 2016-10-27T20:38:39

$[NaCl(aq)]=0.55*mol*L^-1$ .

$"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$ .

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

$"Moles of NaCl(i)"$ $=$ $1.50*dm^3xx0.30*mol*dm^-3=0.45*mol$

$"Moles of NaCl(ii)"$ $=$ $2.50*dm^3xx0.70*mol*dm^-3=1.75*mol$

And thus $[NaCl]$ $=$ $(0.45*mol+1.75*mol)/(4.0*dm^3)$

$=$ $0.55*mol*dm^-3$ .

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

A 1.5*dm^31.5*dm^3 volume of 0.30*mol*dm^-30.30*mol*dm^-3 NaCl(aq)NaCl(aq) was mixed with a 2.5*dm^32.5*dm^3 volume of 0.70*mol*dm^-30.70*mol*dm^-3 NaCl(aq)NaCl(aq). What is the final concentration of NaCl(aq)NaCl(aq)?