Respuesta :
In 250 mL of volumetric flask add 0.975875 grams of [tex]CaF_2[/tex] and dissolve it in the 250 mL of water.
Given:
The solid of calcium fluoride.
To prepare:
The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.
Method:
Molarity of the fluoride ion solution needed = M = 0.100 M
The volume of the fluoride ion solution needed = V = 250 mL
[tex]1 mL = 0.001L\\V=250 mL=250\times 0.001 L=0.250 L[/tex]
The moles of fluoride ion needed = n
According to the definition of molarity:
[tex]M=\frac{n}{V}\\0.100M=\frac{n}{0.250 L}\\n=0.100M\times 0.250 L=0.025 mol[/tex]
Moles of fluoride ion = 0.025 mol
We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.
[tex]CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)[/tex]
According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:
[tex]=\frac{1}{2}\times 0.025 mol=0.0125 \text{mol of } CaF_2[/tex]
Moles of calcium fluoride = 0.0125 mol
Mass of calcium fluoride needed to prepare the solution :
[tex]=0.0125 mol\times 78.07 g/mol=0.975875 g[/tex]
Preparation:
- Weight 0.975875 grams of calcium fluoride
- Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
- Now add a small amount of water to dissolve the calcium fluoride completely.
- After this add more water up to the mark of the volumetric flask of volume 250 mL.
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