Question

For \[{H_2}{O_2}\] solution used for hair bleaching is sold as a solution of approximately \[5.0\,g\,{H_2}{O_2}\] per \[100\,\,ml\] of the solution. The molecular mass of \[{H_2}{O_2}\] is \[34\]. The molarity of this solution is approximately.
A.\[0.15\,M\]
B.\[1.5\,M\]
C.\[3.0\,M\]
D.\[3.4\,M\]

Answer 1

Hint: Here, the molecular mass is given for \[{H_2}{O_2}\] is \[34\]. To calculate the molarity of \[{H_2}{O_2}\] solution, first calculate the number of moles of \[{H_2}{O_2}\] in \[100\,\,ml\] of the solution and then put all values in the formula of molarity.
Formula used:
\[Molarity\, = \,\,\dfrac{{Number\,of\,moles\,of\,solute\,(mol)}}{{Volume\,of\,solution\,\,(liters)}}\]

Complete answer:
We know that, molarity (M) is defined as the ratio of the number of moles of solute to the volume of solution in liters. Its unit is \[moles/liter\]. It is used to measure the concentration of a solution.
In other words, molarity indicates the number of moles of solute per liter of solution. It is also used to calculate the volume of solvent or the amount of solute.
\[Molarity\, = \,\,\dfrac{{Number\,of\,moles\,of\,solute\,(mol)}}{{Volume\,of\,solution\,\,(liters)}}\]
If the solutions has same amount of moles of solute then, it can be represented as \[{C_1}{V_1}\, = \,{C_2}{V_2}\] where \[C\] is the concentration and \[V\] is volume.
According to the question,
Given weight of \[{H_2}{O_2}\]=\[5.0\,g\]
Molecular mass of \[{H_2}{O_2}\]=\[34\,g\,mo{l^{ - 1}}\]
We know that, \[Number\,of\,moles = \dfrac{{Given\,\,weight\,(g)}}{{Molecular\,mass(gmo{l^{ - 1}})}}\]
Number of moles of \[{H_2}{O_2}\]=\[\dfrac{5}{{34}}\,\,mol\]
Number of moles of \[{H_2}{O_2}\]= \[0.147\,mol\]
It is given that,
Volume of \[{H_2}{O_2}\] solution =\[100\,\,ml\]= \[0.1\,L\]
Using the formula for molarity,
\[Molarity\, = \,\,\dfrac{{Number\,of\,moles\,of\,solute\,(mol)}}{{Volume\,of\,solution\,\,(liters)}}\]
Molarity = \[\dfrac{{0.147}}{{0.1}}\]
Molarity =\[1.47\,mol\,{L^{ - 1}}\]\[ \approx 1.5\,mol\,{L^{ - 1}}\]
Hence, the molarity of solution is \[1.5\,mol\,{L^{ - 1}}\]
The correct answer is option (B).

Note:
Don’t get confused between molarity and molality. There is a difference between molarity and molality. Molality is defined as the number of moles of solute dissolved in one Kilogram of solvent. It is denoted by m. Its unit is \[mol/Kg\]. It is also used to determine the concentration of solutions. It gives very precise and accurate values. However, molarity may be imprecise and accurate.