First and Second Moment of Mass I recently came across the definition of the Center of Mass of a system as the point about which the first moment of mass is zero.
Further, it defined Moment of Inertia as the second moment of mass.
My question is, What is this 'moment of mass'?
 A: Given some distribution or density $\rho(x),$ a moment is the 'expectation value' of some power of $x \in \mathbb{R}$. To be precise, the $n$-th moment $M_n$ is given by
$$M_n = \int_{\mathbb{R}} x^n \rho(x) \mathrm{d}x.$$
In the mechanics case, $\rho(x)$ is simply the mass density.
You can extend this to vectors in $\mathbb{R}^d$ in a straightforward way; for example, for the moment of inertia you replace $x^2$ by $\mathbf{x}^2 = x_1^2 + \ldots x_d^2$ to obtain
$$I = M_2 = \int_{\mathbb{R}^d} \mathbf{x}^2 \rho(\mathbf{x}) \mathrm{d}^dx$$
which should match the definition given in your mechanics textbook.
For the first moment of mass, you need to distinguish different directions. As you indicate, you can choose your coordinates such that
$$\int_{\mathbb{R}^d} x_i \,\rho(\mathbf{x}) \mathrm{d}^d x = 0$$
where $i$ runs over the coordinates. In three dimensions, you have $x_1 = x, x_2=y$ and $x_3=z.$
A: 
I don't know whether this is right or wrong, coz its like bringing back the high school...

When physicists define a "moment of something", then it necessarily means Distance $\times$ the "something". Moment of mass simply implies Distance $\times$ Mass.
For a system of $n$ particles, in order to obtain the center of mass - we consider a reference point. The effective mass times the distance to center of mass (which is a moment) will be equal to the sum of moments of individual masses. If $x_{c}$ is the distance from center of mass to the reference point, then
$$\sum_{i=1}^nm_i\ x_{c}=\sum_{i=1}^nm_i x_i$$
Hence, at the center of mass - plugging both equations to the left side,
$$\sum_{i=1}^nm_i(x_{c}-x_i)=0$$
Well, I think this is the first moment of mass (which has also equated to zero).

As far as I can see in your definition, I guess that the second moment roughly says, it is the moment of (moment of mass) which means $$I=\sum_{i=1}^nm_ix_i\times x_i\implies I=MX^2$$ For my luck, it also satisfies with the units $\text{kg m}^2$.
