Answer:
[tex]m_{HNO_3}=283.5gHNO_3[/tex]
Explanation:
Hello,
In this case, since the molarity is defined in terms of the moles of solute and the volume of solution in liters as follows:
[tex]M=\frac{mol}{V}[/tex]
In such a way, solving for the moles of nitric acid, we obtain:
[tex]mol=M*V=1.5L*3mol/L=4.5molHNO_3[/tex]
Finally, since the molar mass of nitric acid is 63 g/mol, the mass is computed as:
[tex]m_{HNO_3}=4.5molHNO_3*\frac{63gHNO_3}{1molHNO_3} \\\\m_{HNO_3}=283.5gHNO_3[/tex]
Regards.
