The atomic mass of element B is 106.1 amu.
Given:
To find:
The atomic mass of element B.
Solution:
The mass of element A = 28.54 g
The atomic mass of element A = 35.45 amu = 35.45 g/mol
Moles of element A :
= [tex]\frac{28.54 g}{35.45 g/mol}=0.8051 mol[/tex]
A +B → AB
According to reaction, one mole of A combines with 1 mole of B , then 0.8051 moles of A will combine with:
[tex]=\frac{1}{1}\times 0.8051 mol=0.8051 \text{ mol of B}[/tex]
Moles of B element = 0.8051 mol
Mass of element B used = 85.42 g
The atomic mass of element B =M
[tex]0.8051 mol=\frac{85.42 g}{M}\\M=\frac{85.42 g}{0.8051 mol}=106.1 g/mol = 106.1 amu[/tex]
The atomic mass of element B is 106.1 amu.
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