Answer:
[tex]\%m/m=4\%[/tex]
Explanation:
Hello,
In this case, by knowing that the molarity is measured in molal units which are mole per liter of solution and the by-mass percentage demands us to compute the mass of the solution, we proceed by assuming 1 L of solution:
[tex]m_{solution}=1L*\frac{1000mL}{1L}*\frac{1.0107g}{1mL} =1010.7g[/tex]
Then, for 1 L of solution, we have 0.756 moles of solute (ammonium chloride), so we compute the grams for those moles by using its molar mass of 53.491 g/mol as shown below:
[tex]m_{solute}=0.756mol*\frac{53.491g}{1mol}=40.4g[/tex]
Finally, we compute the by-mass percentage as shown below:
[tex]\%m/m=\frac{m_{solute}}{m_{solution}} *100\%=\frac{40.4g}{1010.7g}*100 \%\\\\\%m/m=4\%[/tex]
Best regards.