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nevsk [136]
3 years ago
15

A student is instructed to make 1 L of a 2.0 M solution of CaCl2 using dry salt. How should he do this?

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
4 0
<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.
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Calculate the amount of HCl in grams required to react with 3.75 g of CaCO3 according to the following reaction: CaCO3(s) + 2 HC
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