Answer:
- [tex]M=0.38M[/tex]
- [tex]\\ \% m=3.67\%[/tex]
Explanation:
Hello,
In this case, since the molar mass of potassium nitrate is 101.1 g/mol, we can compute the molarity as follows:
[tex]M=\frac{75.1g*\frac{1mol}{101.1g} }{1.95L} \\\\M=0.38M[/tex]
Moreover, as the mass percent is computed as:
[tex]\% m=\frac{m_{KNO_3}}{m_{solution}} *100\%[/tex]
Thus, by using the given density of the solution, we obtain:
[tex]\% m=\frac{75.1g}{1.95L*\frac{1000mL}{1L}*\frac{1.05g}{1mL} } *100\%\\\\ \% m=3.67\%[/tex]
Regards.
