Question

a container holds 3.0 l of 1.0 molar kclo₃ solution. how many moles of clo₃⁻ ions are in the solution? kclo₃ → k⁺ + clo₃⁻ 1.0 mole clo₃⁻ 2.0 moles clo₃⁻ 3.0 moles clo₃⁻

Explanation:

Step1: Recall Molarity Formula

Molarity ($M$) is defined as moles of solute per liter of solution, so $M=\frac{n}{V}$, where $n$ is moles of solute and $V$ is volume in liters. We can rearrange to find $n = M\times V$.

Step2: Calculate Moles of $\ce{KClO3}$

Given $M = 1.0\space M$ (molar) and $V = 3.0\space L$. Plugging into the formula: $n_{\ce{KClO3}}=1.0\space mol/L\times3.0\space L = 3.0\space mol$.

Step3: Relate Moles of $\ce{KClO3}$ to $\ce{ClO3^-}$

From the dissociation equation $\ce{KClO3 -> K+ + ClO3^-}$, the mole ratio of $\ce{KClO3}$ to $\ce{ClO3^-}$ is $1:1$. So moles of $\ce{ClO3^-}$ equal moles of $\ce{KClO3}$, which is $3.0\space mol$.

Answer:

3.0 moles $\ce{ClO3^-}$