Question

What Is The Concentration Of Sugar ( C12H22O11 ) In $\bold{mole \: \: {l}^{ -l} }$ If It's 20 Gram Are Dissolved In Enough Water To Make A Final Volume Up To 2 Litre ?

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Answer 1

The easiest way to answer this question is to first figure out the molar mass of the sugar in question. To do this multiply the number of individual atoms for a given element by its atomic mass.

12 X 12.01 g/mol = 144.12 g/mol C

22 X 1.008 g/mol = 22.176 g/mol H

11 X 16.00 g/mol = 176.00 g/mol O

Total = 342.296 ~ 342.30 g/mol C12H22O11

These numbers can be found using the periodic table

The second step is to take the number of grams of sugar you have; in this case 20g and divide it by the molar mass of the sugar 342.30 g/mol.

20g = 0.058 mol C12H22O11

342.30g/mol

The Third step is to take the number of moles and divide it by your total volume of 2L. Remember that for liquid concentrations, it is always in units of moles/L; this is also the definition of Molarity.

0.058 mol = 0.029 mol/L OR 0.029M C12H22O11

2L

I hope this clears up any confusion

Answer 2

$\bold{Molar \: Mass \: Of \: Sugar \: ( \: C12H22O11 \: ) \: = \: 12 \: × \: 12 + \: 1 \: × \: 22 \: + \: 16 \: × \: 11 \: } \\ \bold{Molar \: Mass \: Of \: Sugar \: ( \: C12H22O11 \: ) \: = \: \: 144 \: + \: 22 \: + \: 176 \: } \\ \bold{Molar \: Mass \: Of \: Sugar \: ( \: C12H22O11 \: ) = \: 342 \: u \: }$

$\fbox{ \fbox{ \bold{ \large{Given : - \: }}}}$

$\bold{Mass \: Of \: The \: Sugar \: = \: 20 \: Gram \: } \\ \\ \bold{Volume \: Of \: The \: Solution \: = \: 2 \: Litre \: } \\ \\ \bold{Molar \: Mass \: Of \: Sugar \: = \: 342 \: u \: }$

$\fbox{ \fbox{\bold{\large{To \: Find \: :- \: }}}}$

${ \bold{\large{Molarity = \:? \: }}}$

$\fbox{ \fbox \bold{\large{ \: Solution \: :- \: }}}$

$\bold{We \: Know \: That \: , \: }$

$\fbox{ \fbox{ \bold{ \huge{Molarity = \: \frac{Mass \: }{Molar \: Mass \: \times \: Volume \: ( \: In \: Litre \: ) \: } }}}}$

$\large{\bold{Molarity \: = \: \frac{20}{342 \times 2} }} \\ \\ \large{\bold{Molarity \: = \: 0.0292 \: \: mole \: {l}^{ - 1} }}$