Question

1. calculate the molarity of a phosphoric acid (h₃po₄) solution that is 84% by mass phosphoric acid and has a density of 1.87 g/ml.

Explanation:

Step1: Assume 1 L of the solution

The density of the solution is $1.87\ g/mL$. For a $V = 1000\ mL$ (1 L) solution, the mass of the solution $m_{solution}$ is calculated using the density formula $
ho=\frac{m}{V}$. So, $m_{solution}=
ho\times V$.
$m_{solution}=1.87\ g/mL\times1000\ mL = 1870\ g$

Step2: Calculate the mass of phosphoric acid

The solution is 84% by mass phosphoric acid. So the mass of $H_3PO_4$, $m_{H_3PO_4}=0.84\times m_{solution}$
$m_{H_3PO_4}=0.84\times1870\ g = 1570.8\ g$

Step3: Calculate the molar mass of phosphoric acid

The molar - mass of $H_3PO_4$: $M_{H_3PO_4}=(3\times1 + 31+4\times16)\ g/mol=98\ g/mol$

Step4: Calculate the number of moles of phosphoric acid

The number of moles $n=\frac{m}{M}$. So, $n_{H_3PO_4}=\frac{m_{H_3PO_4}}{M_{H_3PO_4}}$
$n_{H_3PO_4}=\frac{1570.8\ g}{98\ g/mol}=16.03\ mol$

Step5: Calculate the molarity

Molarity $M=\frac{n}{V}$ (where $V$ is in liters). Since $V = 1\ L$ and $n = 16.03\ mol$, the molarity $M = 16.0\ M$ (rounded to one decimal place)

Answer:

$16.0\ M$