<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is HCl.
are the n-factor, molarity and volume of base which is 
We are given:

Putting values in above equation, we get:

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

Molarity of solution = 0.011 M
Volume of solution = 36.0 mL
Putting values in above equation, we get:

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.
Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles
Hence, the moles of hydroxide ions present in the sample is 0.0008 moles