Answer :
Answer:
[tex]M=0.1584M[/tex]
Explanation:
Hello,
In this case, a mixing process is carried out, so the molarity should be modified by considering the new volume of the solution after mingling, in such a way, the resulting moles are:
[tex]n_{NaCl}=0.031L*0.180molNaCl/L+0.080L*0.150molNaCl/L\\n_{NaCl}=0.01758molNaCl[/tex]
Afterwards, the total new volume of the solution is:
[tex]V_f=31.0mL+80.0mL=111.0mL*\frac{1L}{1000mL}=0.111L[/tex]
Finally, the new molality is:
[tex]M=\frac{0.01758mol}{0.111L}=0.1584M[/tex]
Best regards.