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MrRissso [65]
3 years ago
15

What would be the final volume (in mL) of the aqueous potassium fluoride which was prepared by dissolving 8.0 g of potassium flu

oride in an appropriate volume of water to give a 0.89 M aqueous solution?
Chemistry
1 answer:
andrew11 [14]3 years ago
5 0

n=m/M

8.0g÷58.10g/mol=0.14mol

V=n/c

0.14mol÷0.89M=0.16L=1.6×10²mL

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Determine the concentration of a solution made by dissolving 1.40grams NaCL in enough water to make 30.0mL of solution.
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Explanation:

Given that,

Amount of moles of NaCl (n) = ?

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For molar mass of NaCl, use the molar masses:

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Chlorine, Cl = 35.5g

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Since, amount of moles = mass in grams / molar mass

n = 1.40g / 58.5g/mol

n = 0.024 mole

Now, given that:

Amount of moles of NaCl (n) = 0.024

Volume of NaCl solution (v) = 30.0mL

[Convert 30.0mL to liters

If 1000 mL = 1L

30.0mL = 30.0/1000 = 0.03L]

Concentration of NaCl solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

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c = 0.024 mole / 0.03 L

c = 0.8 M (0.8M means concentration is in moles per litres)

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If HCl is a 0.05 M solution and 50 mL is used to titrate the NaOH, what is the molarity of NaOH if the flask contains 100 mL?
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Now we will put the values:

C₁V₁  = C₂V₂

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C₂  = 2.5 M.mL /100 mL

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