Can their mass being irrational ? or It's integer ?

If I understand the problem correctly, the answer is 2. Let the weight of the real coins be $x$ and that of the fake be $y$. This solution works regardless of whether $x,y$ are specified to be integers or reals. To show 1 use is not enough, suppose our only weighing is with $a_i$ coins taken from pile $i$. This can't distinguish between the cases where $y = x + a_2$ and pile 1 is the only one with fake coins, or where $y = x + a_1$ and pile 2 is the only one with fake coins. For 2 uses of the scale to work, first weigh one coin from each pile. If we get weight $nx$, all coins are real. Otherwise, we get a weight $nx + \delta$ for some error $\delta$ (positive or negative). Then we know $y-x = \delta/j$ for some $j \in \{1,2, \ldots n\}$. Now do a weighing with $k^{i-1}$ coins taken from pile $i$ for $i = 1, 2, \ldots n$, where $k$ is an integer greater than $n^2 + 1$. The total where all coins are real would be $W = \frac{k^n-1}{k-1} x$. If pile $i$ contains the fake coins, this will offset us from $W$ by at least $a_i = k^{i-1}|\delta|/n$ and at most $b_i = k^{i-1}|\delta|$. Now note that \[a_{i+1} = \frac{k^i |\delta|}{n} > n\frac{(k^i - 1) |\delta|}{k-1} = n|\delta| (k^{i-1} + k^{i-2} + \cdots + 1) = n(b_i + b_{i-1} + \cdots + b_1).\] Therefore, if $W'$ is the weight we actually got on the second weighing, we can take the maximum $i$ such that $|W - W'| > a_i$ and know from this that pile $i$ is fake while piles $i+1$ through $n$ are real. Therefore, $|W - W'| = k^i |\delta|/j + S$ ($\ast$) where $S$ is a nonnegative real with $S \leq b_{i-1} + b_{i-2} + \cdots + b_1 < \frac{k^i |\delta|}{n^2}$ ($S$ is the "error" resulting from piles $1$ through $i-1$). Since we know both $|W - W'|$ and $\delta$, and $|k^i \delta/m - k^i \delta/m'| > S$ for any distinct $m,m' \in \{1, 2, \ldots n\}$, there is only one possibility for $j$ that can satisfy ($\ast$) and thus we can find what $y$ is. Determining which piles are which from this is easy.