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Question: The density of water and ethanol at room temperature is \(1.0g/mL\)and \(0.789g/mL\)respectively. Wh...

The density of water and ethanol at room temperature is 1.0g/mL1.0g/mLand 0.789g/mL0.789g/mLrespectively. What volume of ethanol contains the same number of molecules as are present in 100mL100mLof water?
A. 307.8mL307.8mL
B. 168.7mL168.7mL
C. 323.9mL323.9mL
D. 253.3mL253.3mL

Explanation

Solution

As we know, one mole contains 6.022×10236.022 \times {10^{23}}particles of molecules. For example, one mole hydrogen contains 6.022×10236.022 \times {10^{23}}hydrogen particles and similarly one mole oxygen contains 6.022×10236.022 \times {10^{23}}oxygen particles. We have to remember that number of moles (N)=MassMolar  Mass(N) = \dfrac{{Mass}}{{Molar\;Mass}}

Complete step by step answer:
At first, we have to determine the volume which contains the same number of molecules as in 100mL100mLwater. It can be possible only when moles of ethanol and water are equal.
Now, the number of moles (N)=MassMolarMass(N) = \dfrac{{Mass}}{{MolarMass}}
We can write nH2O=nC2H5OHn{H_2}O = n{C_2}{H_5}OH……. (i) as moles of water and ethanol is equal.
Now, we need mass and molar mass of water to determine the moles of water.
Molar mass of water is =(H2O)=2×1+16=18gmol−1 = ({H_2}O) = 2 \times 1 + 16 = 18gmo{l^{ - 1}}
Now, we will calculate the mass of water using the density formula. Given the density of water is 1.0g/mL1.0g/mLand the volume of water is 100mL.100mL.
Density=MassVolumeDensity = \dfrac{{Mass}}{{Volume}}
⇒1.0g/mL=Mass  Of  Water100mL\Rightarrow 1.0g/mL = \dfrac{{Mass\;Of\;Water}}{{100mL}}
⇒MassOfWater=100g\Rightarrow MassOfWater = 100g
Hence, the number of moles of water is,
⇒nH2O=100g18gmol−1\Rightarrow n{H_2}O = \dfrac{{100g}}{{18gmo{l^{ - 1}}}}
Similarly, we can determine the moles of ethanol (C2H5OH)({C_2}{H_5}OH)
Molar mass of ethanol is =(C2H5OH)=2×12+1×6+16=46gmol−1 = ({C_2}{H_5}OH) = 2 \times 12 + 1 \times 6 + 16 = 46gmo{l^{ - 1}}
Given the density of ethanol is 0.789g/mL.0.789g/mL.Consider the volume of ethanol is V.
Density=MassVolumeDensity = \dfrac{{Mass}}{{Volume}}
⇒0.789g/mL=Mass  Of  EthanolV\Rightarrow 0.789g/mL = \dfrac{{Mass\;Of\;Ethanol}}{V}
⇒Mass  Of  Ethanol=0.789g/mL×V\Rightarrow Mass\;Of\;Ethanol = 0.789g/mL \times V
Hence, the number of moles of ethanol is,
⇒nC2H5OH=0.789g/mL×V46gmol−1\Rightarrow n{C_2}{H_5}OH = \dfrac{{0.789g/mL \times V}}{{46gmo{l^{ - 1}}}}
Now, put the number of moles of water and ethanol in equation (i).
nH2O=nC2H5OHn{H_2}O = n{C_2}{H_5}OH
⇒100g18gmol−1=0.789g/mL×V46gmol−1\Rightarrow \dfrac{{100g}}{{18gmo{l^{ - 1}}}} = \dfrac{{0.789g/mL \times V}}{{46gmo{l^{ - 1}}}}
⇒V=100g18gmol−1×0.789g/mL×V46gmol−1\Rightarrow V = \dfrac{{100g}}{{18gmo{l^{ - 1}}}} \times \dfrac{{0.789g/mL \times V}}{{46gmo{l^{ - 1}}}}
⇒V=323.9mL\Rightarrow V = 323.9mL

So, the correct answer is Option C.

Note: The number 6.022×10236.022 \times {10^{23}}is named in honor of the Italian physicist Amedeo Avogadro. The Avogadro’s number is used to count very small particles. The relation between mole and Avogadro’s number is 1mol=6.022×1023particles.1mol = 6.022 \times {10^{23}}particles.