An aqueous solution of calcium hydroxide is standardized by titration with a 0.183 M solution of hydrochloric acid. If 16.6 mL of base are required to neutralize 25.8 mL of the acid, what is the molarity of the calcium hydroxide solution?

0.142 M

Explanation:

Let's consider the neutralization reaction between calcium hydroxide and hydrochloric acid.

Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O

25.8 mL of 0.183 M HCl are used. The reacting moles are:

0.0258 L × 0.183 mol/L = 4.72 × 10⁻³ mol

The molar ratio of Ca(OH)₂ to HCl is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 4.72 × 10⁻³ mol = 2.36 × 10⁻³ mol

2.36 × 10⁻³ moles of Ca(OH)₂ are contained in 16.6 mL. The molarity of Ca(OH)₂ is:

M = 2.36 × 10⁻³ mol/0.0166 L = 0.142 M

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.