We can use the formula for pH to calculate the concentration of hydrogen ions in each solution:

$$\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$$

For the first solution, we have:

$$1 = -\log_{10}[\mathrm{H}^+]$$

Solving for $[\mathrm{H}^+]$, we get:

$$[\mathrm{H}^+] = 0.1 \mathrm{~M}$$

For the second solution, we want:

$$2 = -\log_{10}[\mathrm{H}^+]$$

Solving for $[\mathrm{H}^+]$, we get:

$$[\mathrm{H}^+] = 0.01 \mathrm{~M}$$

To dilute the first solution to the desired concentration, we can use the dilution equation:

$$C_1V_1 = C_2V_2$$

where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume.

We know $C_1 = 0.1 \mathrm{~M}$, $C_2 = 0.01 \mathrm{~M}$, and $V_1 = 1 \mathrm{~L}$. Solving for $V_2$, we get:

$$V_2 = \frac{C_1V_1}{C_2} = \frac{(0.1 \mathrm{~M})(1 \mathrm{~L})}{0.01 \mathrm{~M}} = 10 \mathrm{~L}$$

Therefore, we need to add $10-1=9$ liters of water to the initial solution to obtain the desired pH. Converting liters to milliliters, we get:

$$9 \mathrm{~L} \times \frac{1000 \mathrm{~mL}}{1 \mathrm{~L}} = 9000 \mathrm{~mL}$$

So the volume of water needed is 9000 mL or 9,000 mL (nearest integer).