Question

36. (5 points) The following questions relate to 25.0 grams of Sr(OH)2. a. How many moles of Sr(OH), are present? b. If dissolved in 250 mL of water, what is the [Sr(OH)]?! c. How many Sr(OH), formula units are present? d. How many OH-ions are present?

Answer 1

To solve this we need molar mass of Sr(OH)2
molar mass = [1*72.04 + 2*16 + 2*1.01] g/mol =121.63 g/mol
given mass of Sr(OH)2 = 25 g

SO number of moles present= given mass/molar mass= 25 g/121.63 g/mol
= 0.2055 mols
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given
volume of solution = 250 ml = 0.25 L
number of moles of Sr(OH)2 present = 0.2055 mols
therefore concentration os solution ,
molarity = number of moles of Sr(OH)2 present /volume of solution
= 0.2055 mols/0.250 L = 0.822 M

SO [Sr(OH)2] = 0.822 M
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we have
number of moles of Sr(OH)2 present = 0.2055 mols

1mole of particle contain avogadro number of particles [6.022 x10^23]

Thus 1 mole Sr(OH)2 has 6.022 x10^23 number of formula unit
so 0.2055 mole Sr(OH)2 has 6.022 x10^23*0.2055 number of formula unit
=1.238 x10^23 formula units

Number of formula units =1.238 x10^23 formula units
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we have Sr(OH)2 ---> Sr2+ + 2OH-
in solution the salt ionizes as shown above. 1mole Sr(OH)2 produce 2 mole OH-
Since there are 1.236 x10^23 formula units of Sr(OH)2
number of OH- ions = 2*1.238 x10^23 = 2.476 x10^23 ions
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