Question

how many grams of kmno₄ are needed to make 0.500 l of 1.0 m kmno₄ solution?
kmno₄: 158.03 g/mol
? g

Explanation:

Step1: Recall the formula for molarity

Molarity ($M$) is defined as moles of solute ($n$) divided by volume of solution in liters ($V$), so $M=\frac{n}{V}$. We can rearrange this to find moles: $n = M\times V$.
Given $M = 1.0\ M$ (or $\frac{mol}{L}$) and $V = 0.500\ L$, we calculate moles of $\ce{KMnO4}$:
$n = 1.0\ \frac{mol}{L} \times 0.500\ L = 0.500\ mol$.

Step2: Calculate mass from moles

The molar mass of $\ce{KMnO4}$ is given as $158.03\ g/mol$. The formula relating moles, molar mass ($MM$), and mass ($m$) is $n=\frac{m}{MM}$, so we rearrange to $m = n\times MM$.
Substituting $n = 0.500\ mol$ and $MM = 158.03\ g/mol$:
$m = 0.500\ mol\times158.03\ g/mol = 79.015\ g$.

Answer:

$79.0\ g$ (or $79\ g$ or $79.015\ g$ depending on significant figures, here with three significant figures from $0.500\ L$ and $1.0\ M$, it's $79.0\ g$)