Explanation needed in Central Limit Theorem

I am studying Central Limit Theorem and have seen the equation, $\sqrt{n}(\bar{X}-\mu) \sim N(0, \sigma^2)$, but I am wondering why the $\sqrt{n}$ is multiplied in the equation. At the end of the calculation, the equation becomes $(\bar{X}-\mu) \sim N(0, 1)$. The problem is the lecturer does say why he has put $\sqrt{n}$. Can somebody tell me why I can put the $\sqrt{n}$?

Think about what happens to $\bar X-\mu$ as the sample size $n$ increases. We expect the sample mean to go toward the true mean, right? So if we looked at this quantity, it would converge to zero.
However, if we multiply by $\sqrt{n}$ it turns out that compensates for the above convergence to zero, and the random quantity $\sqrt{n}(\bar X-\mu)$ has a nice standard normal distribution.
The reason the correct quantity is $\sqrt{n}$ is because the variance of $\bar X$ is $$Var(\bar X) = Var\left(\frac{\sum_i X_i}{n}\right) = \frac{1}{n^2}\sum_iVar(X) = \frac{Var(X)}{n}$$ So that $$Var(\sqrt{n}(\bar X-\mu)) = nVar(\bar X) = 1.$$