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Question: What is the molarity of a sucrose solution that contains 10.0g of \({{C}_{12}}{{H}_{22}}{{O}_{11}}\)...

What is the molarity of a sucrose solution that contains 10.0g of C12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}(342.34 g/mol) dissolved in 100.0 mL of solution?

Explanation

Solution

Attempt this question by thinking about the definition of the molarity so as to calculate the molarity of a sucrose solution. As we know that molarity or molar concentration of the solution is a concentration term which is defined as number of moles of solute in per litre of solution.
Formula used:
We will use the following formula for the solution:-
M=nVM=\dfrac{n}{V}
where,
M = Molarity of the solution.
n = number of moles of solute present in the solution.
V = volume of solution in litres.

Complete answer:
Let us first begin with discussion of molarity followed by the calculation as follows:-
Molarity: It is also known as molar concentration which is used to define the number of moles of solute present in per litre of the solution. It can be calculated by using the following formula:-
M=nVM=\dfrac{n}{V}
where,
M = Molarity of the solution.
n = number of moles of solute present in the solution.
V = volume of solution in litres.
-Calculation of moles of sucrose (C12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}):-
Molar mass of C = 12g/mol
Molar mass of H = 1g/mol
Molar mass of O = 16g/mol
Molar mass of C12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}= 12(12g/mol) + 22(1g/mol) + 11(16g/mol) = 342g/mol
Given mass of C12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}= 10.0g
So the number of moles of sucrose (C12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}) = Given mass of C12H22O11Molar mass of C12H22O11\dfrac{\text{Given mass of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}}{\text{Molar mass of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}}
nC12H22O11=10g342g/mol=0.029mol{{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=\dfrac{10g}{342g/mol}=0.029mol
-Calculation of molarity ofC12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}:-
Given that the volume of solution is = 100.0 mL = 0.1L
So the molarity of C12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}= nC12H22O11V\dfrac{{{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}}{V}
0.029mol0.1L=0.29mol/L\Rightarrow \dfrac{0.029mol}{0.1L}=0.29mol/L
-Hence, the molarity of a sucrose solution that contains 10.0g of C12H22O11{{C}_{12}}{{H}_{22}}{{O}_{11}}(342.34 g/mol) dissolved in 100.0 mL of solution is 0.29mol/L.

Note:
-Do not forget to convert the given values into the units required as per the solution so as to obtain an accurate answer and also perform calculations with the units for minimum errors.
-Concentration terms chapter must be read thoroughly as they play an important part in most of the physical chemistry questions.