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nevsk [136]
2 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]2 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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7 0
1 year ago
_____ NaPO4+_____KOH
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NaPO4 + KOH -> KPO4 + NaOH
already balance
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Water is initially present in a state where it's molecules are far apart. During a change of state it's molecules slow down whic
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How many moles of k3po4 can be formed when 4.4 moles of h3po4 react with 3.8 moles of koh
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1.27moles of K3PO4 can be formed.

7 0
3 years ago
Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere
Colt1911 [192]
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
            = (2x10^-16)/(1x10^-7)^2
             = 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
          = 2x10^-8 M
c) at PH= 14 
when POH = 14-PH
                   = 14 - 14 
                   = 0
when POH = -㏒[OH]
              0 = - ㏒[OH]
∴[OH] = 1 m 
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M


8 0
3 years ago
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