The carbonate of a metal is isomorphouswith magnesium carbonate and contains 6.091% carbon the atomic mass of the element is ?
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Let the formula of unknown carbonate = XCO3
Molecular weight of XCO3 = (x +12 + 48)g = (60 + x)g
x is the atomic mass of element X
% composition of carbon = [12/(60 + x)] × 100
6.091= [12/(60 + x)] × 100
0.06091 = [12/(60 +x)]
0.06091 (60 +x) =12
3.6546 + 0.06091x = 12
0.06091x = 12- 3.6546
0.06091x = 8.3454
x = 8.3454/0.06091
x = 137.01
Atomic wt. of the element is 137.01a.m.u which is Barium and the metal carbonate is BaCO3
Molecular weight of XCO3 = (x +12 + 48)g = (60 + x)g
x is the atomic mass of element X
% composition of carbon = [12/(60 + x)] × 100
6.091= [12/(60 + x)] × 100
0.06091 = [12/(60 +x)]
0.06091 (60 +x) =12
3.6546 + 0.06091x = 12
0.06091x = 12- 3.6546
0.06091x = 8.3454
x = 8.3454/0.06091
x = 137.01
Atomic wt. of the element is 137.01a.m.u which is Barium and the metal carbonate is BaCO3
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