The weights of the coins are $2 \text{gm}$, $5 \text{gm}$, $10 \text{gm}$, $25 \text{gm}$, and $50 \text{gm}$.

Statement 1:

Let’s find out the minimum number of coins required to make $78 \text{gm}$. For this we need to maximize the number of heavier coins. So, we will use:

1 coin of $50 \text{gm}$ 2 coins of $10 \text{gm}$ 4 coins of $2 \text{gm}$

So, the minimum number of coins required to buy $78 \text{gm} = 1 + 2 + 4 = 7$

So, Statement 1 is correct.

Statement 2:

To weigh $78$ grams, we can use $80$ grams on one side and $2$ gram coin on the other side.

To make $80$ grams, we need: 1 coin of $50 \text{gm}$ 1 coin of $25 \text{gm}$ 1 coin of $5 \text{gm}$

So, we need 4 coins to weigh $78$ grams – one coin each of $50 \text{gm}$, $25 \text{gm}$, $5 \text{gm}$ and $2 \text{gm}$.

So, Statement 2 is correct.

Hence, option (C) is correct.