Question

Elements A and B form two compounds B2A3 and B2A; 0.05 moles of B2A3 weigh 9.0 gram and 0.10

mole of B2A weighs 10 gm. Atomic weights of A and B are

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Answer 1

B2A3 .....  molar mass =  9g/ 0.05 

                                    =  900/5   = 180

     Therefore relative formula mass of B2A3= 180

that means, (atomic mass A)x 2 + (atomic mass B)x3  =  180

                    2A + 3B = 180

Relative atomic mass of B2A:

molar mass= mass / moles  =    10g/0.1 moles

                                             =    100 u  

            

                   Therefore RAM of B2A= 100

That means,    (atomic mass B)x2 + atomic mass A= 100

                         2B  +  A = 100.....

Solve the two equations by substitution;

B2 + A3 =180

2B + A   = 100  ( A= 100 -2B)

B2 + 3( 100 -2B) = 180

2B + 300 - 6B = 180

2B- 6B = 180-300

 -4B = -120  (divide both sides by -4)

    B  =  30

substitute B to find A

2B + A = 100

(2 x30) + A = 100            60 + A = 100

                                             A = 100 - 60

                                              A = 40

Therefore Atomic weight of A is 40

                                                B is 30