Answer
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Hint: The molarity of a solution is defined as the number of moles of solute dissolved in one litre of solution. Mathematically it is given as:
\[{\text{Molarity}} = \dfrac{{{\text{No}}{\text{.of moles of solute}}}}{{{\text{volume of solution (in ml)}}}} \times 1000\]
Here, No of moles of a compound in a given sample is defined as the weight of the compound divided by the molecular weight or Avogadro’s number i.e., $6.023 \times {10^{23}}$.
Complete step by step answer:
As we know that the density of pure water is 1000kg/m3 and 1Kg is equal to 1000g and 1$m^3$ is equal to 106 $cm^3$. Thus 1000$ Kg/m^3$ = $10g/c{m^3}$. Since, $Density = \dfrac{{Mass}}{{Volume}}$ (1)
As we know molarity means the number of moles in a one-litre solution. Hence the volume of the solution is $1Ltr = 1000c{m^3}$. On putting the value of density and volume in equation (1), we get the value of mass:
$1g/c{m^3} = \dfrac{{Mass}}{{1000c{m^3}}}$
$ \Rightarrow 1 \times 1000 = Mass$
$ \Rightarrow Mass = 1000g$
\[{\text{No}}{\text{.of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\] and the molar mass of water is given as 18g/mol. Hence,
\[{\text{No}}{\text{.of moles}} = \dfrac{{1000}}{{18}} = 55.6M\]
So, the correct answer is Option B.
Additional Knowledge:
Stoichiometry is defined as the relationship between chemical compounds in a reaction. The concepts related to concentration or unit of concentrations are: normality, molarity, molality, percent by volume, percent by weight. The first three are more important and commonly used.
Note: Also, the molarity and molality affects the rate of reaction and are important in reactions involving aqueous solution. The reactions for the process such as precipitation and neutralization, involve the concept of molarity and it is used in stoichiometric calculations for the same. Another concept involving the concentration is molality and it is defined as the one kilogram of solvent for every one litre of solution.
\[{\text{Molarity}} = \dfrac{{{\text{No}}{\text{.of moles of solute}}}}{{{\text{volume of solution (in ml)}}}} \times 1000\]
Here, No of moles of a compound in a given sample is defined as the weight of the compound divided by the molecular weight or Avogadro’s number i.e., $6.023 \times {10^{23}}$.
Complete step by step answer:
As we know that the density of pure water is 1000kg/m3 and 1Kg is equal to 1000g and 1$m^3$ is equal to 106 $cm^3$. Thus 1000$ Kg/m^3$ = $10g/c{m^3}$. Since, $Density = \dfrac{{Mass}}{{Volume}}$ (1)
As we know molarity means the number of moles in a one-litre solution. Hence the volume of the solution is $1Ltr = 1000c{m^3}$. On putting the value of density and volume in equation (1), we get the value of mass:
$1g/c{m^3} = \dfrac{{Mass}}{{1000c{m^3}}}$
$ \Rightarrow 1 \times 1000 = Mass$
$ \Rightarrow Mass = 1000g$
\[{\text{No}}{\text{.of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\] and the molar mass of water is given as 18g/mol. Hence,
\[{\text{No}}{\text{.of moles}} = \dfrac{{1000}}{{18}} = 55.6M\]
So, the correct answer is Option B.
Additional Knowledge:
Stoichiometry is defined as the relationship between chemical compounds in a reaction. The concepts related to concentration or unit of concentrations are: normality, molarity, molality, percent by volume, percent by weight. The first three are more important and commonly used.
Note: Also, the molarity and molality affects the rate of reaction and are important in reactions involving aqueous solution. The reactions for the process such as precipitation and neutralization, involve the concept of molarity and it is used in stoichiometric calculations for the same. Another concept involving the concentration is molality and it is defined as the one kilogram of solvent for every one litre of solution.
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