Please help! show how all was answered and if i filled it in right so far? The d

Question

Please help! show how all was answered and if i filled it in right so far? The density of water 997kg/m^3

a)volume of titrant used in the analysis

b)moles of titrant used in the analysis

c)moles of unknown acid in the analysis

d)molar mass of unknown acid

e)average molar mass

f)standard deviation

g)%RSD

significant figures must be applied accordingly wishes to determine the concentration of an unknown chloride sample for a local water e was found to have a concentration of 0.2086 M. The data collected is summarized rompt A student department. The student was provided a standard solution of silver below nitrate. The standard solution of O.2086M AgNO3(aq) + Cl. (aq) ? AgCl (s) + NOS (1) Trial 3 Trial 1 25.00 0.26 70.65 Trial 2 25.00 0.12 87.55 Student Data Value 25.00 0.05 69.04 Sample Volume (mL) Initial Volume of Titrant (mL): Final Volume of Titrant (mL) Volume of Titrant (mL) Moles of Titrant (mol) Moles of Unknown Acid (mol) Molar Mass of Unknown Acid (g/mol): Average Molar Mass of Unknown Acid (g/mol) Standard Deviation (g/mol) % RSD following values ne of titrant used in the analysis

Explanation / Answer

let us consider ur solution for trial 1

(a) volume of titrant used in analysis = Initial burette reading - final burette reading

=70.65-0.26 =70.39 ml

concentration of titrant = moles of titrant/ volume of solution

moles of titrant = 0.2086* 70.39 =14.683 milimoles = 0.014 moles = moles of unknown acid , since both react in same molar ratio

(b) MIVI= M2V2

0.2086*70.39 = M2 * 25

=0.5873 M

number of moles of unknown acid =0.014

mass of acid = volume of acid * density of water

25* 1 = 25 gm = 0.025 kg

molar mass of acid = 0.025/ 0.014 =1.785 kg= 178.5 gm

* plz specify me what is RSD?

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