Question

Which of the following correctly represents $360\mathrm{g}$ of water?

$$\begin{gathered} \left(\mathrm{i}\right)2\mathrm{moles}\mathrm{of}{\mathrm{H}}_{2}0 \\ \left(\mathrm{ii}\right)20\mathrm{moles}\mathrm{of}\mathrm{water} \\ \left(\mathrm{iii}\right)6.022\times {10}^{23}\mathrm{molecules}\mathrm{of}\mathrm{water} \\ \left(\mathrm{iv}\right)1.2044\times {10}^{25}\mathrm{molecules}\mathrm{of}\mathrm{water} \end{gathered}$$

Choose the option :

• A. $\left(\mathrm{i}\right)$
• B. $\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{iv}\right)$
• C. $\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right)$
• D. $\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iv}\right)$

Answer 1

The correct option is D

$\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iv}\right)$

Explanation for correct option

(d) $\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iv}\right)$

• When a substance is of one mole, it is equal to $6.022\times {10}^{23}$unit of that substance.
• This unit is known as Avogadro's number.
• Step $1$: Find number of moles of given mass of water

$\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{water}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{water}}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{water}}$

• Here, $$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{water}\mathrm{given}=360\mathrm{g} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{water}=18\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$
• Hence, $$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{360\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}} \\ =20\mathrm{mol} \end{gathered}$$
• Step $2$: Find number of molecules of given mass of water
• $$\begin{gathered} \mathrm{number}\mathrm{of}\mathrm{molecules}=\mathrm{Number}\mathrm{of}\mathrm{moles}\times \mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{number} \\ =20\times 6.022\times {10}^{23} \\ =1.2044\times {10}^{25}\mathrm{molecules}\mathrm{of}\mathrm{water} \end{gathered}$$
• Hence given mass of water has $20\mathrm{moles}$ and $1.2044\times {10}^{25}\mathrm{molecules}$ of water present.
• Therefore, the correct option is (d) $\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iv}\right)$

Explanation for incorrect options

After calculations, it was found that the given mass of water has $20\mathrm{moles}$ and $1.2044\times {10}^{25}\mathrm{molecules}$ of water present. Hence, option (a), (b) and (c) are incorrect.

Hence, the correct option is (d) $\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iv}\right)$.