3a) How many mL of a 1.21 M Na
3a) How many mL of a 1.21 M NaOH solution would contain 0.275moles of NaOH? ___ mL NaOH solution
3b) How many mL of 12M HCl solution is needed to prepare 775mLof 0.880M HCl by dilution? ___ mL is needed.
3c) Calculate the molarity of a solution made by dissolving35.2g K2CO3 in enough water to form 275mL of solution. The solutionis ____ 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 a0.33M K2CO3 solution? ____ M
Answer:
Answer – 3a) Given, [NaOH] = 1.21 M , moles ofNaOH = 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 toprepare 775mL of 0.880M HCl by dilution
3c) mass of K2CO3 = 35.2g , volume = 275 mL = 0.275 L
First we need to calculate the moles ofK2CO3
Moles of K2CO3 = 35.2 g / 138.205g.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] = 0.33M
We know, 1 [K2CO3] = 2*[K+]
So, [K+] =2*0.33
= 0.66 M
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