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The number of hydrogen atoms present in $25.6$ $\mathrm{g}$ of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ which has a molar mass of $342.3 \mathrm{~g}$ is
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The correct answer is:
$9.91 \times 10^{23}$
No. of moles of a compound
$=\frac{\text { given mass }(\mathrm{gm})}{\text { Molar mass }(\mathrm{gm})}$
i.e. $\frac{25.6}{342.3}=0.0748$ moles.
1 mole of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ contains $6.022$ $\times 10^{23}$ molecules of it.
Hence $0.0748$ moles contains
$\begin{array}{l}
=6.022 \times 10^{23} \times 0.0748 \\
=0.4504 \times 10^{23} \text { molecules. }
\end{array}$
1 molecule of sucrose by formula is having 22 atoms of hydrogen.
$\therefore 0.4504 \times 10^{23} \times 10^{23} \times 22=9.91 \times 10^{23}$ atoms of hydrogen.
$=\frac{\text { given mass }(\mathrm{gm})}{\text { Molar mass }(\mathrm{gm})}$
i.e. $\frac{25.6}{342.3}=0.0748$ moles.
1 mole of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ contains $6.022$ $\times 10^{23}$ molecules of it.
Hence $0.0748$ moles contains
$\begin{array}{l}
=6.022 \times 10^{23} \times 0.0748 \\
=0.4504 \times 10^{23} \text { molecules. }
\end{array}$
1 molecule of sucrose by formula is having 22 atoms of hydrogen.
$\therefore 0.4504 \times 10^{23} \times 10^{23} \times 22=9.91 \times 10^{23}$ atoms of hydrogen.
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