I did the following question but I\'m not sure with my work and wantverification

Question

I did the following question but I'm not sure with my work and wantverification on my answer...thanks

Question 1: Show how to use dimensional analysis to answer thefollowing.

(a) What is the molarity of a 100 mL solution containing 0.35 g ofpotassium sulfate?



(b) How many moles of sodium chloride are present in 1.00 L of a0.150 M NaCl solution?



(c) Concentrated ammonia, NH3, is 14.7M. How many mL ofthis solution is needed to make up 2.0L of a 1.0 M NH3,solution?


(d) Hydrochloric acid reacts with potassium carbonate. Provide thebalanced chemical equation showing the reactants and products. Howmany mL of 0.15 M potassium carbonate will react with 25mL of 1.00M HCl?

Explanation / Answer

molar mass of potassium sulfate,K2SO4   = 174.23g/ mol So 0.35g of potassium sulfate = 0.35g / 174.23g/mol =0.002011 molarity = moles / volume = 0.002011mol / 100mL =0.02011M ------------- moles = molarity * volume            =0.150M * 1.0L = 0.150mol/L * 1.0L = 0.150mol mass = moles * molar mass         = 0.150mol * 58.44g/mol = 8.766g ---------------- McVc = MdVd 14.7M * Vc = 1.0M * 2.0L Vc =  1.0M * 2.0L / 14.7M =0.1361L = 136.1mL ---------------- 2HCl (aq) + K2CO3 (aq)...............> KCl (aq) + H2CO3(aq) Since these two are reacts in 2 : 1 ratio ; atequilivalence point moles of HCl = moles ofK2CO3 (25.0mL * 1.0M ) /2 = 0.15M *VK2CO3 VK2CO3 ={(25.0mL * 1.0M) / 2 *0.15M } / = 83.33mL VK2CO3 ={(25.0mL * 1.0M) / 2 *0.15M } / = 83.33mL

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