Question

What NaCl concentration results when 284 mL of a 0.850 M NaCl solution is mixed with 467 mL of a 0.380 M NaCl solution

Answer 1

Answer:

0.558 M.

Explanation:

Hello!

In this case, when computing the resulting concentration of the solution after a mixing process, we first need to compute the moles provided by each solution:

$n_{NaCl}^{sol1}=0.284L*0.850mol/L=0.2414mol\\\\n_{NaCl}^{sol2}=0.467L*0.380mol/L=0.1775mol$

So the total number of moles are:

$n_T=0.2414mol+0.1775mol=0.4189mol$

And the total volume which is additive:

$V_T=0.284L+0.467L=0.751L$

Therefore, the resulting molarity of sodium chloride is:

$M=\frac{0.4189mol}{0.751L}\\\\M=0.558M$

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