Question

a 500. ml aqueous solution of na₃po₄ (molar mass = 164 g/mol) was prepared using 82 g of the solute. what is the molarity of na₃po₄ in the resulting solution?
a 0.0010 m
b 0.16 m
c 0.25 m
d 1.0 m

Explanation:

Step1: Calculate moles of solute

Use the formula $n=\frac{m}{M}$, where $m = 82\ g$ (mass of solute) and $M=164\ g/mol$ (molar - mass of solute).
$n=\frac{82\ g}{164\ g/mol}=0.5\ mol$

Step2: Convert volume to liters

Given volume $V = 500\ mL$. Since $1\ L=1000\ mL$, then $V=\frac{500}{1000}L = 0.5\ L$

Step3: Calculate molarity

Use the molarity formula $M=\frac{n}{V}$, where $n$ is the number of moles of solute and $V$ is the volume of the solution in liters.
$M=\frac{0.5\ mol}{0.5\ L}=1.0\ M$

Answer:

D. 1.0 M