Answer:
C. 1.07 M.
Explanation:
Hello,
In this case, we can define the molarity of the bleach as shown below:
[tex]M=\frac{moles_{NaClO}}{V_{solution}}[/tex]
In such a way, given the mass of bleach in a 1-L solution, we can compute the density:
[tex]\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL[/tex]
In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:
[tex]V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L[/tex]
Afterwards, using the by-mass percent of bleach we compute the mass:
[tex]m_{NaClO}=100g*0.0725=7.25g[/tex]
And the moles:
[tex]n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol[/tex]
Therefore, the molarity turns out:
[tex]M=\frac{0.097mol}{0.091L}\\ \\M=1.07M[/tex]
Thus, answer is C. 1.07 M.
Regards.
