given that the molar mass of nano3 is 85.00 g/mol, what mass of nano3 is needed to make 4.50 l of a 1.50 m…

given that the molar mass of nano3 is 85.00 g/mol, what mass of nano3 is needed to make 4.50 l of a 1.50 m nano3 solution? use molarity = moles of solute / liters of solution. 6.75 g 18.9 g 255 g 574 g

given that the molar mass of nano3 is 85.00 g/mol, what mass of nano3 is needed to make 4.50 l of a 1.50 m nano3 solution? use molarity = moles of solute / liters of solution. 6.75 g 18.9 g 255 g 574 g

Answer

Answer:

C. 255 g

Explanation:

Step1: Calculate moles of solute

We know that molarity $M=\frac{n}{V}$, where $M$ is molarity, $n$ is moles of solute and $V$ is volume of solution in liters. Rearranging for $n$, we get $n = M\times V$. Given $M = 1.50\ mol/L$ and $V=4.50\ L$, so $n=1.50\ mol/L\times4.50\ L = 6.75\ mol$.

Step2: Calculate mass of solute

We know that $m = n\times M_m$, where $m$ is mass, $n$ is moles and $M_m$ is molar - mass. Given $n = 6.75\ mol$ and $M_m=85.00\ g/mol$, so $m=6.75\ mol\times85.00\ g/mol=573.75\ g\approx574\ g$.