Suppose that A &B form the compounds B2A3 and B2A. If 0.05 moles of B2A3 weighs 9g and 0.1 mole of B2A weighs 10g. The atomic weight of A and B respectively are
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To find the atomic weight for the elements A and B, we will use the following formulas:
Moles = mass/ molar mass.....molar mass= mass/moles
Hence you can calculate the molar mass of the compounds B2A3 and B2A using the above formulas.
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
Hope you find this clear and helpful.
Moles = mass/ molar mass.....molar mass= mass/moles
Hence you can calculate the molar mass of the compounds B2A3 and B2A using the above formulas.
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
Hope you find this clear and helpful.
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Heya mate,
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Here is your answer...
Kindly check the attachment✌️
Hope it helps❤️❣️
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