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Question: In which case is the number of molecules of water maximum? A. \({10^{ - 3}}\) mol of water. B. 0...

In which case is the number of molecules of water maximum?
A. 10−3{10^{ - 3}} mol of water.
B. 0.00224 L of water vapour at 1 atm and 273 k.
C. 0.18 g of water.
D. 18 ml of water.

Explanation

Solution

To solve this question, we have to remember that the number of molecules can be calculated by the formula, number of molecules = number of moles ×\times NA{N_A}, where NA{N_A} = Avogadro number = 6.022×10236.022 \times {10^{23}} and the number of moles can be calculated by using the formula, number of moles = given massmolar mass\dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}

Complete step by step answer:
First, we have to find out the molar mass of water.
So, molecular formula of water = H2O{H_2}O
Molecular mass = 2(molar mass of hydrogen)+molar mass of oxygen2\left( {{\text{molar mass of hydrogen}}} \right) + {\text{molar mass of oxygen}}
Molecular mass = 2(1)+162\left( 1 \right) + 16
Molecular mass = 18 g.
Therefore, 1 mole of water = 18 g of water.
Let us find the number of molecules of water for each option.

(A). 10−3{10^{ - 3}} mol of water.
Given that, Number of moles = 10−3{10^{ - 3}}
Number of molecules = number of moles ×\times 6.022×10236.022 \times {10^{23}},
Putting the value, we will get
Number of molecules = 10−3×6.022×1023{10^{ - 3}} \times 6.022 \times {10^{23}}
Number of molecules = 6.022×10206.022 \times {10^{20}}

(B). 0.00224 L of water vapour at 1 atm and 273 k.
We know that, according to the ideal gas equation
⇒n=pVRT\Rightarrow n = \dfrac{{pV}}{{RT}}, where n is the number of moles, p is the pressure, V is the volume, T is the temperature and R is the gas constant, R = 0.08210.0821L atm K−1mol−1{K^{ - 1}}mo{l^{ - 1}}
Here we have, V = 0.00224 L, p = 1 atm and T = 273 K
Putting all these values in above formula, we will get
⇒n=1×0.002240.0821×273\Rightarrow n = \dfrac{{1 \times 0.00224}}{{0.0821 \times 273}}
⇒n=10−4\Rightarrow n = {10^{ - 4}}moles.
So,
Number of molecules = 10−4×6.022×1023{10^{ - 4}} \times 6.022 \times {10^{23}}
Number of molecules = 6.022×10196.022 \times {10^{19}}

(C). 0.18 g of water.
We know that,
Number of moles = given massmolar mass\dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}
Putting the values,
Number of moles = 0.1818=10−2\dfrac{{0.18}}{{18}} = {10^{ - 2}}moles.
Then,
Number of molecules = 10−2×6.022×1023{10^{ - 2}} \times 6.022 \times {10^{23}}
Number of molecules = 6.022×10216.022 \times {10^{21}}

(D). 18 ml of water.
18 ml of water = 18 g of water. [ mass = volume×\timesdensity, common density of water = 1 gm/ml ]
Number of moles = 1 mole
So, Number of molecules = 1×6.022×10231 \times 6.022 \times {10^{23}}
Number of molecules = 6.022×10236.022 \times {10^{23}}
Hence, from all the values, we can see that the number of molecules of water is maximum 6.022×10236.022 \times {10^{23}} in 18ml of water.
So, the correct answer is “Option D”.

Note: Whenever we ask such a question, we should know some basic concepts of chemistry. In case of gases, a mole is defined as the amount of the gas which occupies a volume of 22.4 L at STP, which is called its molar volume and the mass of one mole of substance in gram is called its molar mass.