3a) How many mL of a 1.21 M NaOH solution would contain 0.275 moles of NaOH? ___

ID: 941164 • Letter: 3

Question

3a) How many mL of a 1.21 M NaOH solution would contain 0.275 moles of NaOH? ___ mL NaOH solution

3b) How many mL of 12M HCl solution is needed to prepare 775mL of 0.880M HCl by dilution? ___ mL is needed.

3c) Calculate the molarity of a solution made by dissolving 35.2g K2CO3 in enough water to form 275mL of solution. The solution is ____ M.

3d) 10.5mL of 12M HCl solution is diluted to a volume of 122mL. What is the concentration of the resulting solution? ___M

3e) What is the molar concentration of potassium ions (K+) in a 0.33M K2CO3 solution? ____ M

Answer – 3a) Given, [NaOH] = 1.21 M , moles of NaOH = 0.275 moles

We know the ,

Molarity = moles / L

So, volume (L) = moles / molarity

= 0.275 moles / 1.21 M

= 0.227 L

= 227 mL

3b) Given, M1 = 12 M , V1 = ? , M2 = 0.880 M , V2 = 775 mL

M1V1 = M2V2

So, V1 = M2*V2/M1

= 0.880 M * 775 mL / 12 M

= 56.83 mL

So, 56.8 mL of 12M HCl solution is needed to prepare 775mL of 0.880M HCl by dilution

3c) mass of K2CO3 = 35.2 g , volume = 275 mL = 0.275 L

First we need to calculate the moles of K2CO3

Moles of K2CO3 = 35.2 g / 138.205 g.mol-1

= 0.255 moles

So, molarity of K2CO3 = 0.255 moles / 0.275 L

= 0.926 M

3d) Given, M1 = 12 M , V1 = 10.5 , M2 = ? M , V2 = 122 mL

M1V1 = M2V2

So, M2 = M1*12/V2

= 12.0 M * 10.5 mL / 122 M

= 1.03 M

3e) Given, [K2CO3] = 0.33 M

We know, 1 [K2CO3] = 2 *[K+]

So, [K+] = 2*0.33

= 0.66 M

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