Answer:
(a) ![102.6g/mol](https://tex.z-dn.net/?f=102.6g%2Fmol)
(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:
![V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}](https://tex.z-dn.net/?f=V_%7Bbase%7D%3D0.1L%3B%20M_%7Bacid%7D%3D2.5M%2C%20V_%7Bacid%7D%3D0.017L%5C%5CV_%7Bbase%7DM_%7Bbase%7D%3DV_%7Bacid%7DM_%7Bacid%7D)
Solving for the molarity of base we've got:
![M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L](https://tex.z-dn.net/?f=M_%7Bbase%7D%3D%5Cfrac%7BM_%7Bacid%7D%2AV_%7Bacid%7D%7D%7BV_%7Bbase%7D%7D%3D%5Cfrac%7B2.50M%2A0.017L%7D%7B0.1L%7D%20%3D0.425M%3D0.425mol%2FL)
Now, we can compute the moles of the base as:
![n_{base}=0.425mol/L*0.1L=0.0425mol](https://tex.z-dn.net/?f=n_%7Bbase%7D%3D0.425mol%2FL%2A0.1L%3D0.0425mol)
(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:
![\frac{4.36g}{0.0425mol} =102.6g/mol](https://tex.z-dn.net/?f=%5Cfrac%7B4.36g%7D%7B0.0425mol%7D%20%3D102.6g%2Fmol)
(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:
![102.6g/mol-16g/mol-1g/mol=85.6g/mol](https://tex.z-dn.net/?f=102.6g%2Fmol-16g%2Fmol-1g%2Fmol%3D85.6g%2Fmol)
So, that atomic mass dovetails to the Rubidium's atomic mass.
Best regards.