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Phoenix [80]
4 years ago
10

You have equal masses of different solutes dissolved in equal volumes of solution. Which of the solutes would make the solution

having the highest molar concentration?
a. LiOH
b. KCl
c. KOH
d. NaOH
e. All are the same
Chemistry
2 answers:
mrs_skeptik [129]4 years ago
6 0

Answer:

a. LiOH

Explanation:

The relationship between the parameters (mass, volume and molar concentration) mentioned in the question is given as;

Molar Concentration = (Mass / Molar Mass) * Volume

Upon calculating thr molar mass of all the compounds we have;

a)  LiOH - 6.94 + 16 + 1.01 = 24 g/mol

b) KCl - 39.098 + 35.45 = 74.5 g/mol

c) KOH - 39.098 + 16 + 1.01 = 56.1 g/mol

d) NaOH - 28.9898 + 16 + 1.01 = 46 g/mol

Picking 1 g and 1 L as the values for mass and volume. Any value would work, so long as it is constant for all compounds.

a) 1 g * 1 mol/24 g * 1/1L = 0.042 mol/L

b) 1 g * 1 mol/74.5 g * 1/1L = 0.013 mol/L

c) 1 g * 1 mol/56.1 g * 1/1L = 0.018 mol/L

d) 1 g * 1 mol/46 g * 1/1L =0.021 mol/L

LiOH has the highest molar concentration.  

Aleonysh [2.5K]4 years ago
3 0

Answer:

The highest molar concentration is in the solution of potassium chloride because it's the one, with the highest molar mass.

Explanation:

If we have the same mass and the same volume in all the solutes and solutions, we should know the molar mass for each, to calculate how many moles of solute are we having.

a. LiOH → 23.94 g/m

b. KCl → 74.55 g/m

c. KOH → 56.1 g/m

d. NaOH → 40 g/m

The highest molar mass is in option b. It will have the most moles.

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1214 mg

Explanation:

Data given

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Molarity of NaOH solution (M) = 0.100 M NaOH.

Amount of acetylsalicylic acid = ?

Molecular weight = 180.2 g/mol

Solution:

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As equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.

First find the moles of NaOH

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                 M = moles of solute / 1 L x volume of solution

Rearrange and modify the above equation

                 moles of NaOH = M of NaOH x 1 L / volume of NaOH solution

Put values in above equation

                moles of NaOH = 0.100 M x 1 L / 14.85 mL

                moles of NaOH = 0.0067 moles

Now find to find the weight of acetylsalicylic acid

As

equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.

                 moles of acetylsalicylic acid =  moles of NaOH.......... (1)

As we know  

                  no. of moles = mass in g / molar mass

So,

The modified form of equation 1 will be

              mass in g / molar mass (acetylsalicylic acid) =  moles of NaOH

Put values in above equation

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Rearrange the above equation

                mass of acetylsalicylic acid = 0.0067 moles x 180.2 g/mol

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Convert the grams to milligrams

1 g = 1000 milligram

So,  

1.214 g = 1214 mg

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