Answer
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Hint: We need to remember that when the molarity of a given arrangement is characterized as the complete number of moles of solute per liter of arrangement. The molality of an answer is reliant upon the progressions in actual properties of the framework, for example, pressing factor and temperature as not at all like mass, the volume of the framework changes with the adjustment of states of being of the framework.
Complete answer:
We have to know that the molarity is addressed by M, which is named as molar. One molar is the molarity of an answer where one gram of solute is broken up in a liter of arrangement. As we probably are aware, in an answer, the dissolvable and solute mix to shape an answer, consequently, the all-out volume of the arrangement is taken.
The condition for computing molarity is the proportion of the moles of solute whose molarity is to be determined and the volume of dissolvable used to break down the given solute.
$M = \dfrac{n}{V}$
Where, $M$ is the molarity, $n$ is the no. of moles, and $V$ is the volume of solution.
In the given details, the mass of pure water is $1Kg$ or $1000gm$ and the volume of pure water is $1L$ . First, we have to calculate the number of moles of pure water by using the following expression,
$No.{\text{ of moles = }}\dfrac{{Weight{\text{ of pure water}}}}{{Gram{\text{ of molar mass}}}}$
Where,
The weight of the pure water is $1000gm$ and the molecular mass of the water is $18g/mol$ . Applying all the values in the above expression,
$No.{\text{ of moles = }}\dfrac{{1000g}}{{18g/mol}}{\text{ = 55}}{\text{.56 mol}}$
Then we have to calculate the molarity, by using the following expression,
$Molarity = \dfrac{{No.{\text{ of moles of solute}}}}{{Volume{\text{ of the solution (L)}}}} = \dfrac{{55.56mol}}{{1L}}$
Therefore,
The molarity of pure water is $55.56mol/L$ .
Therefore, the option ( $A$ ) is correct.
Note:
We have to change the unit of density of water from $g/mL$ to $kg/L$ to calculate the molarity of pure water. This is because the molarity is calculated per liter of the solution and we have to make sure that the units are the same.
Complete answer:
We have to know that the molarity is addressed by M, which is named as molar. One molar is the molarity of an answer where one gram of solute is broken up in a liter of arrangement. As we probably are aware, in an answer, the dissolvable and solute mix to shape an answer, consequently, the all-out volume of the arrangement is taken.
The condition for computing molarity is the proportion of the moles of solute whose molarity is to be determined and the volume of dissolvable used to break down the given solute.
$M = \dfrac{n}{V}$
Where, $M$ is the molarity, $n$ is the no. of moles, and $V$ is the volume of solution.
In the given details, the mass of pure water is $1Kg$ or $1000gm$ and the volume of pure water is $1L$ . First, we have to calculate the number of moles of pure water by using the following expression,
$No.{\text{ of moles = }}\dfrac{{Weight{\text{ of pure water}}}}{{Gram{\text{ of molar mass}}}}$
Where,
The weight of the pure water is $1000gm$ and the molecular mass of the water is $18g/mol$ . Applying all the values in the above expression,
$No.{\text{ of moles = }}\dfrac{{1000g}}{{18g/mol}}{\text{ = 55}}{\text{.56 mol}}$
Then we have to calculate the molarity, by using the following expression,
$Molarity = \dfrac{{No.{\text{ of moles of solute}}}}{{Volume{\text{ of the solution (L)}}}} = \dfrac{{55.56mol}}{{1L}}$
Therefore,
The molarity of pure water is $55.56mol/L$ .
Therefore, the option ( $A$ ) is correct.
Note:
We have to change the unit of density of water from $g/mL$ to $kg/L$ to calculate the molarity of pure water. This is because the molarity is calculated per liter of the solution and we have to make sure that the units are the same.
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