A cell with osmolarity measuring 300 mosm/l is placed in a beaker that contains a nacl solution with osmolarity of 150mosm/l. pr
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The question is incomplete, the complete question is:
The solubility of slaked lime, , in water is 0.185 g/100 ml. You will need to calculate the volume of
M HCl needed to neutralize 14.5 mL of a saturated
<u>Answer:</u> The volume of HCl required is 290mL, the mass of is 0.0268g, the moles of
<u>Explanation:</u>
Given values:
Solubility of = 0.185 g/100 mL
Volume of = 14.5 mL
Using unitary method:
In 100 mL, the mass of present is 0.185 g
So, in 14.5mL. the mass of present will be =
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(1)
Given mass of = 0.0268 g
Molar mass of = 74 g/mol
Plugging values in equation 1:
Moles of present =
The chemical equation for the neutralization of calcium hydroxide and HCl follows:
By the stoichiometry of the reaction:
Moles of = Moles of
= 0.000724 mol
The formula used to calculate molarity:
.....(2)
Moles of HCl = 0.000724 mol
Molarity of HCl =
Putting values in equation 2, we get:
Hence, the volume of HCl required is 290mL, the mass of is 0.0268g, the moles of