I honestly don’t know sorry
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
The experimental mole ratio of silver chloride to barium chloride is calculated as below
fin the mole of each compound
mole= mass/molar mass
moles of AgCl = 14.5g/142.5 g/mol = 0.102 moles of AgCl
moles of BaCl2 = 10.2 g/208 g/mol = 0.049 moles of BaCl2
find the mole ratio by dividing each mole with the smallest mole(0.049)
AgCl= 0.102/0.049 =2
BaCl2 = 0.049/0.049 =1
therefore the mole ratio AgCl to BaCl2 is 2 :1
The answer to your question is a 12
