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Answer to Question #11960 in Inorganic Chemistry for leah
Question #11960
calculate the mass of each of the following
1) 1.95x 10^22 molecules of sucrose, C12 H22 O11
2) 0.780 mol Ca(CN)2
3) 9.57 X 10^17 atoms platinum
1) 1.95x 10^22 molecules of sucrose, C12 H22 O11
2) 0.780 mol Ca(CN)2
3) 9.57 X 10^17 atoms platinum
Expert's answer
1) The sucrose molar mas is 342.3 g/mol. Amount of sucrose is
n(C12 H22 O11)
= 1.95E22 / 6.022E23 = 0.032 mol
The mass of sucrose is
m(C12 H22 O11) =
n(C12 H22 O11) * M(C12 H22 O11) = 0.032 mol * 342.3 g/mol = 10.95 g
2)
The molar mass of Ca(CN)2 is 92.1 g/mol.
The mass of Ca(CN)2
is
m(Ca(CN)2) = n(Ca(CN)2) * M(Ca(CN)2) = 0.78 mol * 92.1 g/mol = 71.84
g
3) The platinum molar mas is 195 g/mol. Amount of platinum is
n(Pt)
= 9.57E17 / 6.022E23 = 1.6E-6 mol = 0.0000016 mol
The mass of platinum
is
m(Pt) = n(Pt) * M(Pt) = 1.6E-6 mol mol * 195 g/mol = 0.0003 g = 0.3 mg
n(C12 H22 O11)
= 1.95E22 / 6.022E23 = 0.032 mol
The mass of sucrose is
m(C12 H22 O11) =
n(C12 H22 O11) * M(C12 H22 O11) = 0.032 mol * 342.3 g/mol = 10.95 g
2)
The molar mass of Ca(CN)2 is 92.1 g/mol.
The mass of Ca(CN)2
is
m(Ca(CN)2) = n(Ca(CN)2) * M(Ca(CN)2) = 0.78 mol * 92.1 g/mol = 71.84
g
3) The platinum molar mas is 195 g/mol. Amount of platinum is
n(Pt)
= 9.57E17 / 6.022E23 = 1.6E-6 mol = 0.0000016 mol
The mass of platinum
is
m(Pt) = n(Pt) * M(Pt) = 1.6E-6 mol mol * 195 g/mol = 0.0003 g = 0.3 mg
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