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 = \\frac{moles of solute}{liters of solution}.\n6.75 g\n18.9 g\n255 g\n574 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 = \\frac{moles of solute}{liters of solution}.\n6.75 g\n18.9 g\n255 g\n574 g

Answer

Explanation:

Step1: Calculate moles of solute

We know that molarity $M=\frac{n}{V}$, where $n$ is moles of solute and $V$ is volume of solution in liters. Rearranging for $n$, we get $n = M\times V$. Given $M = 1.50\ M$ 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 mass $m=n\times M_m$, where $M_m$ is molar - mass. Given $n = 6.75\ mol$ and $M_m = 85.00\ g/mol$, so $m=6.75\ mol\times85.00\ g/mol=573.75\ g\approx574\ g$.

Answer:

D. 574 g