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 Molarity Formula

Molarity ($M$) is defined as moles of solute ($n$) per liter of solution ($V$), so $M = \frac{n}{V}$. We need to find moles of $\text{KMnO}_4$ first. Rearranging the formula: $n = M \times V$.
Given $M = 1.0\ \text{M}$, $V = 0.500\ \text{L}$, so $n = 1.0\ \text{mol/L} \times 0.500\ \text{L} = 0.500\ \text{mol}$.

Step2: Calculate Mass from Moles

Mass ($m$) is moles ($n$) times molar mass ($MM$), so $m = n \times MM$.
Given $MM = 158.03\ \text{g/mol}$, $n = 0.500\ \text{mol}$, so $m = 0.500\ \text{mol} \times 158.03\ \text{g/mol} = 79.015\ \text{g}$.

Answer:

79.015 (or approximately 79.0)