
What is the concentration of sugar \[{{\rm{C}}_{12}}{{\rm{H}}_{22}}{{\rm{O}}_{11}}\] in \[{\rm{mol }}{{\rm{L}}^{ - 1}}\] if its 20 g are dissolved in enough water to make a final volume up to 2 L.
Answer
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Hint: To determine the molarity of sugar, use the formula \[{\rm{Molarity}} = \dfrac{{{\rm{Mass of sugar}}}}{{{\rm{Molecular weight of sugar }} \times {\rm{ volume of water}}}}\].
Complete step by step answer:
Molarity is the number of moles of a solute dissolved in one litre of a solution. Number of moles is the ratio of the mass of solute to the molar mass of the solute.
Mass of sugar is 20 grams and the volume of the solution is 2 L.
The molecular formula of the sugar is \[{{\rm{C}}_{12}}{{\rm{H}}_{22}}{{\rm{O}}_{11}}\]. This sugar is cane sugar, also known as sucrose. The atomic weights of carbon, hydrogen and oxygen are 12 g/mol, 1 g/mol and 11 g/mol respectively. Calculate the molecular weight of sugar by adding individual atomic weights of each element.
\[12\left( {{\rm{12 g/mol}}} \right) + 22\left( {{\rm{1 g/mol}}} \right) + 11\left( {{\rm{16 g/mol}}} \right) = 342{\rm{ g/mol}}\]
Hence, the molecular weight of sugar is 342 g/mol.
To determine the molarity of sugar, use the formula
\[{\rm{Molarity}} = \dfrac{{{\rm{Mass of sugar}}}}{{{\rm{Molecular weight of sugar }} \times {\rm{ volume of water}}}}a\]
To calculate molarity, the mass of sugar is divided with the product of molecular weight of sugar and the volume of water.
Substitute values in the above formula and calculate the molarity (concentration) of sugar.
\[\begin{array}{c}
{\rm{Molarity}} = \dfrac{{{\rm{Mass of sugar}}}}{{{\rm{Molecular weight of sugar }} \times {\rm{ volume of water}}}}\\
{\rm{Molarity}} = \dfrac{{{\rm{20 g}}}}{{{\rm{342 g/mol }} \times {\rm{ 2 L}}}}\\
{\rm{Molarity}} = 0.0292{\rm{ mol }}{{\rm{L}}^{ - 1}}
\end{array}\]
Hence the molarity of the solution is \[0.0292{\rm{ mol }}{{\rm{L}}^{ - 1}}\].
Note:
Do not get confused between molarity and molality. Molarity is the number of moles of solute present in one litre of solution. Whereas molality is the number of moles of solute present in one kilogram of solvent.
Complete step by step answer:
Molarity is the number of moles of a solute dissolved in one litre of a solution. Number of moles is the ratio of the mass of solute to the molar mass of the solute.
Mass of sugar is 20 grams and the volume of the solution is 2 L.
The molecular formula of the sugar is \[{{\rm{C}}_{12}}{{\rm{H}}_{22}}{{\rm{O}}_{11}}\]. This sugar is cane sugar, also known as sucrose. The atomic weights of carbon, hydrogen and oxygen are 12 g/mol, 1 g/mol and 11 g/mol respectively. Calculate the molecular weight of sugar by adding individual atomic weights of each element.
\[12\left( {{\rm{12 g/mol}}} \right) + 22\left( {{\rm{1 g/mol}}} \right) + 11\left( {{\rm{16 g/mol}}} \right) = 342{\rm{ g/mol}}\]
Hence, the molecular weight of sugar is 342 g/mol.
To determine the molarity of sugar, use the formula
\[{\rm{Molarity}} = \dfrac{{{\rm{Mass of sugar}}}}{{{\rm{Molecular weight of sugar }} \times {\rm{ volume of water}}}}a\]
To calculate molarity, the mass of sugar is divided with the product of molecular weight of sugar and the volume of water.
Substitute values in the above formula and calculate the molarity (concentration) of sugar.
\[\begin{array}{c}
{\rm{Molarity}} = \dfrac{{{\rm{Mass of sugar}}}}{{{\rm{Molecular weight of sugar }} \times {\rm{ volume of water}}}}\\
{\rm{Molarity}} = \dfrac{{{\rm{20 g}}}}{{{\rm{342 g/mol }} \times {\rm{ 2 L}}}}\\
{\rm{Molarity}} = 0.0292{\rm{ mol }}{{\rm{L}}^{ - 1}}
\end{array}\]
Hence the molarity of the solution is \[0.0292{\rm{ mol }}{{\rm{L}}^{ - 1}}\].
Note:
Do not get confused between molarity and molality. Molarity is the number of moles of solute present in one litre of solution. Whereas molality is the number of moles of solute present in one kilogram of solvent.
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What is the concentration of sugar \[{{\rm{C}}_{12}}{{\rm{H}}_{22}}{{\rm{O}}_{11}}\] in \[{\rm{mol }}{{\rm{L}}^{ - 1}}\] if its 20 g are dissolved in enough water to make a final volume up to 2 L.

Some Basic Concepts of Chemistry | NCERT EXERCISE 1.11 | Class 11 Chemistry Chapter 1 | Nandini Mam
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