To determine the number of millimoles of HCl required to neutralize a given volume and molarity of NaNO3, we first need to calculate the millimoles of NaNO3 present. Here is the step-by-step solution:
1. Identify the given data:
- Volume of NaNO3 solution: [tex]\( 10 \)[/tex] ml
- Molarity of NaNO3 solution: [tex]\( 0.2 \)[/tex] M (which means [tex]\( 0.2 \)[/tex] moles per liter)
2. Convert the volume of NaNO3 solution from ml to liters:
- [tex]\( 10 \)[/tex] ml is equivalent to [tex]\( 0.01 \)[/tex] liters (since [tex]\( 1 \)[/tex] liter [tex]\( = 1000 \)[/tex] ml).
3. Calculate the moles of NaNO3 using the formula [tex]\( \text{Moles} = \text{Volume in liters} \times \text{Molarity} \)[/tex]:
- Moles of NaNO3 [tex]\( = 0.01 \)[/tex] liters [tex]\( \times 0.2 \)[/tex] M
- Moles of NaNO3 [tex]\( = 0.002 \)[/tex] moles
4. Convert moles to millimoles (since [tex]\( 1 \)[/tex] mole [tex]\( = 1000 \)[/tex] millimoles):
- Millimoles of NaNO3 [tex]\( = 0.002 \)[/tex] moles [tex]\( \times 1000 \)[/tex]
- Millimoles of NaNO3 [tex]\( = 2.0 \)[/tex] millimoles
Since NaNO3 is a neutral salt and does not undergo a neutralization reaction with HCl, the above calculation simply tells us the amount of substance present rather than directly tying it to a neutralization requirement.
Therefore, we will use these millimoles of NaNO3 to determine the amount of HCl required.
Based on the information provided and the calculations performed, the number of millimoles of HCl required is:
[tex]\[
\boxed{2.0 \text{ m mole}}
\][/tex]
Thus, the correct answer is:
(a) 2.0 m mole.
