Answer:
(a) [tex]102.6g/mol[/tex]
(b) Rubidium
Explanation:
Hello,
This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:
[tex]V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}[/tex]
Solving for the molarity of base we've got:
[tex]M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L[/tex]
Now, we can compute the moles of the base as:
[tex]n_{base}=0.425mol/L*0.1L=0.0425mol[/tex]
(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:
[tex]\frac{4.36g}{0.0425mol} =102.6g/mol[/tex]
(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:
[tex]102.6g/mol-16g/mol-1g/mol=85.6g/mol[/tex]
So, that atomic mass dovetails to the Rubidium's atomic mass.
Best regards.
