Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

THoneymoon x Ga Blackboard L x D bboard2010 x E-Mail Accesx E Sign into yo x C C

ID: 976867 • Letter: T

Question

THoneymoon x Ga Blackboard L x D bboard2010 x E-Mail Accesx E Sign into yo x C Chemistry q My GRE My x GILove Kanye x CHEM 1215 X /ibiscms/mod/ibis/view.php?id-2501029 WWW sapling learning.com Sapling Learning Sapling Learning My Assignment Resources 4/25/2016 11:55 PM 71.9/100 Gradebook O Assignment Information Attempts Score Print HCalculator Periodic Table Available From: 4/10/2016 06.00 PM 4/25/2016 11:55 PM Question 8 of 8 Points Possible: 100 Map Dab Grade Category: Graded The flask shown here contains 1.04 g of acid and a few drops of phenolphthalein indicator dissolved in water. The buret You can check your answers. contains 0.200 M NaOH. You can view solutions when you complete or give up on any question. What volume of base is needed to You can keep trying to answer each question until reach the end point of the titration? Number in your question for each incorrect attempt at that mL base O Help With This Topic Click to begin Assuming the acid is monoprotic, Web Help & Videos what is its molar mass? Add base to the solution until O Technical Support and Bug Reports it just turns pink. You may Number need to reset the titration if g mol you go past the end point. Previous 3 Give Up & View Solution Check Answer Next Exit careers partner prhag polly terms ofuse contact us m Cortana. Ask me anything Logout & 4x 46 PM 4/25/2016

Explanation / Answer

given data :

mass of an acid = 1.04 g
concentration of NaOH = 0.200 M
The reaction between Base and Monoprotic acid is

NaOH + HX ---> NaX + H2O, as the acid is monoprotic.

Using the above equation, we know that NaOH reacts with HX in a 1 : 1 ratio,

suppose the monoprotic acid is taken as Hcl , then

36.5 grams of Hcl ----------> 40 gram of NaoH

1.04 grams of Hcl -----------> ?

40X1.04 /36.5 = 1.139 grams

Molarity = wt /Gmwt X1000/v

0.2 = 1.139/40 X1000/v

Volume = 142.3 ml

therefore, n (NaOH) = n (HX)

mole= concentration XVolume =0.143 X 0.200 M = 0.0286mol.
Molar Mass of an (acid) = mass of an (acid) / moles of an (acid)

=1.040 g / 0.0286mol

   = 36.36g/mol