find the no. of atoms of each type present in 3.42 g of sugar cane C12H22O11.
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Answer:
The number of carbon, hydrogen and oxygen atoms in 3.42 g of C_{12}H_{22}O_{11}C
12
H
22
O
11
is, 7.226\times 10^{22},1.324\times 10^{23}\text{ and }6.624\times 10^{22}7.226×10
22
,1.324×10
23
and 6.624×10
22
Solution : Given,
Mass of C_{12}H_{22}O_{11}C
12
H
22
O
11
= 3.42 g
Molar mass of C_{12}H_{22}O_{11}C
12
H
22
O
11
= 342 g/mole
First we have to calculate the moles of C_{12}H_{22}O_{11}C
12
H
22
O
11
.
\text{Moles of }C_{12}H_{22}O_{11}=\frac{\text{Mass of }C_{12}H_{22}O_{11}}{\text{Molar mass of }C_{12}H_{22}O_{11}}=\frac{3.42g}{342g/mole}=0.01molesMoles of C
12
H
22
O
11
=
Molar mass of C
12
H
22
O
11
Mass of C
12
H
22
O
11
=
342g/mole
3.42g
=0.01moles
Now we have to calculate the number of atoms of carbon, hydrogen and oxygen.
In the given molecule, there are 12 number of carbon atoms, 22 number of hydrogen atoms and 11 number of oxygen atoms.
As, 1 mole of carbon contains 12\times 6.022\times 10^{23}12×6.022×10
23
number of hydrogen atoms
So, 0.01 mole of carbon contains 0.01\times 12\times 6.022\times 10^{23}=7.226\times 10^{22}0.01×12×6.022×10
23
=7.226×10
22
number of carbon atoms
and,
As, 1 mole of hydrogen contains 22\times 6.022\times 10^{23}22×6.022×10
23
number of hydrogen atoms
So, 0.01 mole of hydrogen contains 0.01\times 22\times 6.022\times 10^{23}=1.324\times 10^{23}0.01×22×6.022×10
23
=1.324×10
23
number of hydrogen atoms
and,
As, 1 mole of oxygen contains 11\times 6.022\times 10^{23}11×6.022×10
23
number of hydrogen atoms
So, 0.01 mole of oxygen contains 0.01\times 11\times 6.022\times 10^{23}=6.624\times 10^{22}0.01×11×6.022×10
23
=6.624×10
22
number of oxygen atoms
Therefore, the number of carbon, hydrogen and oxygen atoms in 3.42 g of C_{12}H_{22}O_{11}C
12
H
22
O
11
is, 7.226\times 10^{22},1.324\times 10^{23}\text{ and }6.624\times 10^{22}7.226×10
22
,1.324×10
23
and 6.624×10
22
Explanation:
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