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Question: What is the concentration of sugar\[\left( {{C_{12}}{H_{22}}{O_{11}}} \right)\] in \[mol{L^{ - 1}}\]...

What is the concentration of sugar(C12H22O11)\left( {{C_{12}}{H_{22}}{O_{11}}} \right) in molL1mol{L^{ - 1}} if it’s 20g20g are dissolved in enough water to make a final volume up to 2L2L?

Explanation

Solution

As we know that the concentration of the sugar can be calculated by the molarity. Because, the molarity is considered as the concentration of the sugar. And the molarity is also known as molar concentration. The molarity states that, the number of moles dissolved in the one liter of the solvent. Hence, it is found by dividing given mass with the product of molar mass and the volume.

Complete answer:
The molarity is also known as molar concentration. So the concentration of sugar can be find out by using the formula of molarity or molar concentration. The chemical equation of molarity can be written as,
molarity=massmolarmass×volumemolarity = \dfrac{{mass}}{{molar mass \times volume}} ……… (1)
Here, the given mass of sugar is equal to20g20g and the volume is equal to 2L2L. The molar mass of sugar is equal to 342g/mol342g/mol.
The calculation to find out the molar mass of sugar is,
C12H22O11=(12×12)+(22×1)+(11×16)=342{C_{12}}{H_{22}}{O_{11}} = \left( {12 \times 12} \right) + \left( {22 \times 1} \right) + \left( {11 \times 16} \right) = 342
Hence, substitute the given values in the above equation, will get
The concentration of sugar, \left( {{C_{12}}{H_{22}}{O_{11}}} \right)$$$$ = \dfrac{{20}}{{342 \times 2}}
On simplification we get,
=0.029mol/L= 0.029mol/L
Therefore, the concentration of the sugar is equal to 0.029mol/L0.029mol/L.

Note:
We need to remember that the molarity or molar concentration is the amount of substance present in a certain volume of the solution. Hence, the molarity is equal to the molar concentration of the solution. Thus, it is also stated that the molarity is the number of moles of solute present in the one liter of the solution. But the molality states that the number of solutes present in the one kilogram of the solvent.