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Question: The density of water and ethanol at room temperature is 1.0 g/mL and 0.789 g/mL respectively. What v...

The density of water and ethanol at room temperature is 1.0 g/mL and 0.789 g/mL respectively. What volume of ethanol contains the same number of molecules as are present in 100 mL of water?
A. 307.8 mL
B. 168.7 mL
C. 323.9 mL
D. 253.3 mL

Explanation

Solution

We know, a mole of anything contains 6.022×10236.022 \times {10^{23}} particles of molecules. For example, a mole of oxygen contains 6.022×10236.022 \times {10^{23}} oxygen particles, a mole of hydrogen contains 6.022×10236.022 \times {10^{23}} oxygen particles. Here, we need the formula of number of moles, Number  of  moles=MassMolarMass{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar Mass}}}}

Complete step by step answer: Here, we have to calculate the volume which contains the same number of molecules as in 100 mL water. Same number of molecules present only when moles of ethanol and water are equal.
The formula to calculate number of moles is,
Number  of  moles=MassMolarMass{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar Mass}}}}
As moles of water and ethanol is equal, we can write
nH2O=nC2H5OH{n_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = {n_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}}
…… (1)
Now, we have to calculate the moles of water. For this, we need mass and molar mass of water. Molar mass of water is=(H2O)=2×1+16=18  g  mol−1\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right) = 2 \times 1 + 16 = 18\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}
Now, we have to calculate the mass of water using the density formula. The given density and volume of water is 1.0 g/mL and 100 mL.
Density=MassVolume{\rm{Density}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Volume}}}}
⇒1.0  g/mL=Mass  of  water100  mL\Rightarrow 1.0\;{\rm{g/mL}} = \dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{water}}}}{{100\;{\rm{mL}}}}
⇒Mass  of  water=100  g\Rightarrow {\rm{Mass}}\;{\rm{of}}\;{\rm{water}} = {\rm{100}}\;{\rm{g}}
So, the moles of water is,
⇒nH2O=100 g18  g mol−1\Rightarrow {n_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = \dfrac{{100\,{\rm{g}}}}{{18\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}}
Similarly we have to calculate the moles of ethanol(C2H5OH)\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}} \right)
The molar mass of ethanol is=2×12+1×6+16=46  g  mol−1 = 2 \times 12 + 1 \times 6 + 16 = 46\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}
The density of ethanol is 0.789 g/mL. Consider the volume of ethanol as V. Now, we use the formula of density.
Mass=0.789   g/mL×V{\rm{Mass}} = 0.789\;\,{\rm{g/mL}} \times V
So, the number of moles of ethanol is,
⇒nC2H5OH=0.789  g/mL×V46 g mol−1\Rightarrow {n_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}} = \dfrac{{0.789\;g/{\rm{mL}} \times V}}{{46\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}}
Now, put the moles of water and ethanol in equation (1).
nH2O=nC2H5OH{n_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = {n_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}}
⇒100 g18  g mol−1=0.789  g/mL×V46 g mol−1\Rightarrow \dfrac{{100\,{\rm{g}}}}{{18\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}} = \dfrac{{0.789\;g/{\rm{mL}} \times V}}{{46\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}}
⇒V=100 g18  g mol−1×46  gmol−10.789 g  /mL\Rightarrow V = \dfrac{{100\,{\rm{g}}}}{{18\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}} \times \dfrac{{{\rm{46}}\;{\rm{g mo}}{{\rm{l}}^{ - 1}}}}{{0.789\,{\rm{g}}\;/{\rm{mL}}}}
⇒323.9  mL\Rightarrow 323.9\;{\rm{mL}}
Therefore, the volume of ethanol is 323.9 mL. Hence, option C is the correct answer.

Note: The number 6.022×10236.022 \times {10^{23}} is named in honor of the Italian physicist Amedeo Avogadro. The Avogadro's number aids in counting very small particles. The relation of mole and Avogadro’s number is 1  mol=6.022×1023  particles1\;{\rm{mol}} = 6.022 \times {10^{23}}\;{\rm{particles}}