
The boiling point of \[{D_2}O\] greater than \[{H_2}O\]. Because
A.\[{D_2}O\] has lower \[{K_w}\] value
B.\[{D_2}O\] has lower dielectric constant
C.\[{D_2}O\] is an associated liquid
D.The molecular weight of \[{D_2}O\] is greater than \[{H_2}O\]
Answer
438.3k+ views
Hint: We have to know that the water is an inorganic compound which does not have any taste and odor. And the water is a volatile liquid which is transparent in nature. The particles present in the liquid are far apart. The \[{D_2}O\] is also known as heavy water and it consists of deuterium rather than hydrogen. And the heavy water is mainly used as a moderator in nuclear reactors and it is also used in NMR-spectroscopy.
Complete answer:
The higher boiling point of heavy water is not due to the lower \[{K_w}\] value of \[{D_2}O\]. Hence, option (A) is incorrect.
The boiling point of \[{D_2}O\] greater than \[{H_2}O\]. But it is not because of the lower dielectric constant of heavy water. Hence, option (B) is incorrect.
The heavy water is an associated liquid. But this is not the reason behind its higher boiling point. Hence, option (C) is incorrect.
The boiling point of heavy water is greater than normal water. Because, the molecular weight of heavy water is greater than the normal water. The molecular weight will affect the boiling point of a liquid.
The molecular weight of \[{D_2}O = \left( {16 + 2 + 2} \right) = 20\]
The molecular weight of \[{H_2}O = \left( {16 + 1 + 1} \right) = 18\]
And heavy water has a higher boiling point. Because, the dispersion force increases with the number of electrons, mass and number of atoms.
Hence, option (D) is correct.
Note:
We need to know that the boiling point is the temperature required to change the phases from liquid state to vapour state. Here, the vapour phase of the liquid becomes equal to the pressure surrounding the liquid. And the liquid is transformed into the vapour state. There are some factors that will affect the boiling point like, atmospheric pressure, molecular weight of liquid, temperature, vapour pressure of liquid etc. And the intermolecular forces present in the liquid also affect the boiling point.
Complete answer:
The higher boiling point of heavy water is not due to the lower \[{K_w}\] value of \[{D_2}O\]. Hence, option (A) is incorrect.
The boiling point of \[{D_2}O\] greater than \[{H_2}O\]. But it is not because of the lower dielectric constant of heavy water. Hence, option (B) is incorrect.
The heavy water is an associated liquid. But this is not the reason behind its higher boiling point. Hence, option (C) is incorrect.
The boiling point of heavy water is greater than normal water. Because, the molecular weight of heavy water is greater than the normal water. The molecular weight will affect the boiling point of a liquid.
The molecular weight of \[{D_2}O = \left( {16 + 2 + 2} \right) = 20\]
The molecular weight of \[{H_2}O = \left( {16 + 1 + 1} \right) = 18\]
And heavy water has a higher boiling point. Because, the dispersion force increases with the number of electrons, mass and number of atoms.
Hence, option (D) is correct.
Note:
We need to know that the boiling point is the temperature required to change the phases from liquid state to vapour state. Here, the vapour phase of the liquid becomes equal to the pressure surrounding the liquid. And the liquid is transformed into the vapour state. There are some factors that will affect the boiling point like, atmospheric pressure, molecular weight of liquid, temperature, vapour pressure of liquid etc. And the intermolecular forces present in the liquid also affect the boiling point.
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