A 35 gram of saturated solution is distilled out completely. If the solubility of the solute is 110, find the amount of residue
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Molecular weight of vanillin = 152 g/mol
Futher molecular formula of vanillin is C8H8O3
Atomic weight of oxygen = 16 g/mol
Thus, 152 g of vanillin contains 16 g of oxygen
∴ 0.045 g (45 mg) of vanillin contains
= 0.00473 g
Also, number of moles of vanillin in 0.045 g sample =
Now, 1 mole = 6.023 X 10^23 molecules
∴ 2.96 X 10^-4 mole = 1.78 X 10^20 molecules
From molecular formula, it can be seen that 1 molecule of vanallin contain 3 atoms of oxygen
∴1.78 X 10^20 molecules contain 3 X 1.78 X 10^20 = 5.34 X 10^20 oxygen atoms
Futher molecular formula of vanillin is C8H8O3
Atomic weight of oxygen = 16 g/mol
Thus, 152 g of vanillin contains 16 g of oxygen
∴ 0.045 g (45 mg) of vanillin contains
Also, number of moles of vanillin in 0.045 g sample =
Now, 1 mole = 6.023 X 10^23 molecules
∴ 2.96 X 10^-4 mole = 1.78 X 10^20 molecules
From molecular formula, it can be seen that 1 molecule of vanallin contain 3 atoms of oxygen
∴1.78 X 10^20 molecules contain 3 X 1.78 X 10^20 = 5.34 X 10^20 oxygen atoms