
A person adds $1.71grams$ of sugar $({C_{12}}{H_{22}}{O_{11}})$ in order to sweeten his tea. The number of carbon atoms added is (Mol. mass of sugar = 342):
A.$3.6 \times {10^{22}}$
B.$7.2 \times {10^{21}}$
C.$0.05$
D.$6.6 \times {10^{22}}$
Answer
550.2k+ views
Hint:As indicated by mole concept one mole of atoms = subatomic mass in grams and number of atoms = number of moles $ \times $ Avogadro's number and so the quantity of atoms is given by the equation = number of moles of substance $ \times $ avo. Number.
Complete answer:
We know that the quantity of moles is the proportion of mass to molar mass.
Then the moles of sugar is equal to $\dfrac{{1.71}}{{342}} = \dfrac{1}{{200}}mol$
Now, we need to calculate the number of molecules ${N_A}$ and this is given by the product of number of moles and Avogadro’s number as:
${N_A} = 6.023 \times {10^{23}}$
Hence, the number of molecules is equal to $\dfrac{1}{{200}} \times {N_A}$
Now let us calculate the number of C atoms in the molecule, and this can be done by using:
Number of C atoms in the molecule = $12 \times \dfrac{1}{{200}} \times {N_A}$
As we know the value of ${N_A}$ let us substitute that in the above and then the value becomes:
$ \Rightarrow 3.6 \times {10^{22}}$
Thus, the correct answer to the above question is option A.
Note:
Where n is the quantity of moles of the substance (or rudimentary element), $N$ is the complete number of rudimentary elements in the example, and ${N_A}$is the Avogadro steady. "Mole" was presented around the year 1896 by the German scientist Wilhelm Ostwald, who got the term from the Latin word moles meaning a 'store' or 'heap.
Complete answer:
We know that the quantity of moles is the proportion of mass to molar mass.
Then the moles of sugar is equal to $\dfrac{{1.71}}{{342}} = \dfrac{1}{{200}}mol$
Now, we need to calculate the number of molecules ${N_A}$ and this is given by the product of number of moles and Avogadro’s number as:
${N_A} = 6.023 \times {10^{23}}$
Hence, the number of molecules is equal to $\dfrac{1}{{200}} \times {N_A}$
Now let us calculate the number of C atoms in the molecule, and this can be done by using:
Number of C atoms in the molecule = $12 \times \dfrac{1}{{200}} \times {N_A}$
As we know the value of ${N_A}$ let us substitute that in the above and then the value becomes:
$ \Rightarrow 3.6 \times {10^{22}}$
Thus, the correct answer to the above question is option A.
Note:
Where n is the quantity of moles of the substance (or rudimentary element), $N$ is the complete number of rudimentary elements in the example, and ${N_A}$is the Avogadro steady. "Mole" was presented around the year 1896 by the German scientist Wilhelm Ostwald, who got the term from the Latin word moles meaning a 'store' or 'heap.
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