1.80g of glucose is dissolved in 36.00g of water in a beaker. The total number of oxygen atoms in the solution is?
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Answered by
26
Number of atoms of oxygen in solution
= Number of atoms of oxygen from glucose (C6H12O6)
+ Number of atoms of oxygen from water
= 1.8/180×6×6.022×10²³+ 36/18×6.022×10²³
= 6.022×10²³(0.18×6/18+36×1/18)
= 6.022×10²³(37.08/18)
= 12.405 × 10²³
= Number of atoms of oxygen from glucose (C6H12O6)
+ Number of atoms of oxygen from water
= 1.8/180×6×6.022×10²³+ 36/18×6.022×10²³
= 6.022×10²³(0.18×6/18+36×1/18)
= 6.022×10²³(37.08/18)
= 12.405 × 10²³
Answered by
31
Hi friend,
Solution :-
In 180gm of glucose(C6H12O6) number of oxigen atom is 6 NA.
In 1.80gm of glucose number of oxigen atom is 6NA/180 × 1.8 = 0.06 NA.
In 18gm of Water the no. of oxigen atom are NA.
In 36.0gm of water the no. of oxigen atom are = Na/18 × 36 = 2 NA.
The total number of oxigen atom is in the solution = (0.06 + 2 ) NA = 2.06NA.
= 2.06 NA × 6.022×10^23
= 12.405 × 10^23...i hope it helps you
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Solution :-
In 180gm of glucose(C6H12O6) number of oxigen atom is 6 NA.
In 1.80gm of glucose number of oxigen atom is 6NA/180 × 1.8 = 0.06 NA.
In 18gm of Water the no. of oxigen atom are NA.
In 36.0gm of water the no. of oxigen atom are = Na/18 × 36 = 2 NA.
The total number of oxigen atom is in the solution = (0.06 + 2 ) NA = 2.06NA.
= 2.06 NA × 6.022×10^23
= 12.405 × 10^23...i hope it helps you
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