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Which solution listed below yields the highest molar concentration?

a. No right answer.

b. 121.45 g of KOH in 75.0 mL

c. 23.49 g of NH4OH in 125.0 mL

d. 217.5 g of LiNO3 in 2.00 L

e. 15.25 g of Pb(C2H3O2)2 in 65.0 mL

User Onika
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1 Answer

3 votes

Answer:

The answer to your question is: KOH

Step-by-step explanation:

Formula

Molarity = # of moles / volume

Process

a. No right answer.

b. 121.45 g of KOH in 75.0 mL

MW KOH = 56 g

56 g --------------------- 1 mol

121.45 g ----------------- x

x = (121.45 x 1) / 56

x = 2.17 mol

M = 2.17 / 0.075

M = 29

c. 23.49 g of NH4OH in 125.0 mL

MW NH4OH = 35 g

35 g -------------------- 1 mol

23.49 g --------------------- x

x = (23.49 x 1) / 35 = 0.67 mol

M = 0.67 / 0.125

M = 5.36

d. 217.5 g of LiNO3 in 2.00 L

MW LiNO3 = 69 g

69 g ---------------------- 1 mol

217.5 g ---------------------- x

x = (217.5 x 1) / 69 = 3.15 mol

M = 3.15 / 2

M = 1.6

e. 15.25 g of Pb(C2H3O2)2 in 65.0 mL

MW Pb(C2H3O2) = 266 g

266 g ------------------ 1 mol

15.25g ................... x

x = (15.25 x 1) / 266 = 0.06 mol

M = 0.06 / 0.065

M = 0.92

User Ishaan
by
5.5k points