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18 - Mole Concept and Stoicheometry Questions Answers

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry IIT JEE

Q. 1.Calculate the number of HCl molecules present in 10ml of 0.1N HCl solution.

Q.2. 4.0g of NaOH are contained in one decilitre of solutions. Calculate mole fraction. Molarity and molality of NaOH. ( Given, density of solution=                                                              1.038g/ cm3)

Q. 3. You are provided with 2 litre each of 0.5 M NaOH and 0.4M of NaOH solutions. What maximum volume of 0.30 M NaOH can be obtained from these solutions without using water at all ?

 

Asked By: HEMA RANI
is this question helpfull: 8 2 read solutions ( 3 ) | submit your answer
Joshi sir comment

1)   no of miligm eq. = 10*0.1 = 1

so no. of gm. eq. = 1/1000

since acid is monobasic so no. of moles = 1/1000

so molecules = 1/1000 * 6.023 * 1023

 

2)     0.1 litre soln. contains              4 gm. NaOH

       so 1 litre contains                      40 gm. NaOH

       so 1 litre soln. contains              40/40 moles of NaOH  = 1 mole so molarity = 1

 

 now 1 litre soln. contains                       40 gm NaOH

so 1*1038 gm soln. contains               40 gm. NaOH                   here 1038 gm/litre is density

so 1038 gm. soln. contains                  40 gm. NaOH

so 1038-40 gm. solvent contains        40 gm. NaOH

so 998 gm. solvent contains                 40 gm. NaOH = 1 mole 

so 0.998 Kg solvent contains                 1 mole NaOH

so molality = 1/0.998 

 

998 gm water contains                   40 gm. NaOH

so 998/18 mole water contains    40/40 mole NaOH

so mole fraction of NaOH = [40/40]/{[40/40]+[998/18]}

 

Third part is irrelevent without using water

 

 

 

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry Medical Exam

IS BASICITY OF ACETIC ACID IS 1? PLS SEND THE BASICITY OF OTHER ACIDS WHICH ARE USUALLY ASKED  IN ENTRANCE EXAM

Asked By: SARIKA SHARMA
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Joshi sir comment

HCl, HNO3 , HCOOH are mono basic acids

COOH-COOH, H2SO4 are di basic acids

similarly others 

 

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry Medical Exam

0.2 g OF SAMPLE OF H2O2  REQUIRED 10ml OF 1N KMnO4 SOLn  IN A TITRATION IN +NCE OF dil H2SO4 . THE % PURITY OF H2O2 IS?

ANS 85

 

Asked By: SARIKA SHARMA
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Joshi sir comment

No. of gm. equivalent of KMnO4 = No. of gm. equivalent of H2O2

so 1*10/1000   =     x  /  17             (17 is eq. wt. of H2O2) 

so  x = 0.17

so % purity = 0.17*100/0.2 = 85 

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry Medical Exam

HOW MUCH WATER IS TO BE ADDED TO DILUTE 10ml OF 10N HCl TO MAKE IT DECINORMAL ?

ANS 100ml

Asked By: SARIKA SHARMA
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Joshi sir comment

N1V1 = N2V2

so 10*10 = 0.1*v 

or v = 1000 ml

so extra water added will be 990 ml

 

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry Medical Exam

the normality of 10% (w/v) acetic acid is ?

1.7N

Asked By: SARIKA SHARMA
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Joshi sir comment

10 % w/v acetic acid means 10 gm. in 100 ml. soln.

so 1000 ml soln. contains 100 gm acetic acid

so 1 litre contains 100/60 moles of acetic acid

so M = 100/60 = 1.67

so N = M * basicity = 1.67*1 = 1.67 

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry Medical Exam

Q ONE LITRE MIXTURE OF CO &CO2 IS PASSED THRU RED HOT CHARCOAL IN A TUBE THE NEW VOL . BECOMES 1.2L WHAT % OF CO IN MIXTURE INITIALLY?

Q AN Aq SOLn CONTAINING 100 g OF DISSOLVED MgSO4 IS FED TO A CRYSTALLIZER WHERE 80% SALT CRY. OUT AS MgSO4.6H2O CRYSTAL HOW MANY gm OF HEXAHYDRATE SALT CRY. ARE OBTAINED FROM CRYSTALLIZER=?

ELECTROCHEM

Fe3+  +  e- -----> Fe2+ = 0.77V

Fe --------.>  Fe3+  + 3e-     E = +0.04V

WHAT IS VALUE OF E* for Fe2+  +2e--------> Fe ? 

Asked By: SARIKA
is this question helpfull: 3 0 submit your answer
Joshi sir comment

CO2 + C (charcoal) ------->  2CO

let us consider that initially volume of CO2 is x litre and that of  CO is (1-x) litre

so after the above written process net CO in final = 2x+1-x = 1.2      given 

on solving x = 0.2 and 1-x = 0.8   now calculate %

 

MgSO4  -------------->   MgSO4 .6H2O

it means 120 gm (mw of compound) gives 228 gm MgSO4 . 6H2O 

so 80 gm will give 228*80/120     solve it

here 80 is used for 80% of 100 gm.

 

 

Fe+++  +  e-  ----->  Fe++                  E=  +0.77 V

Fe                ----->  Fe+++  + 3e-       E = +0.04 V

 

on adding we get

Fe                ----->  Fe++  + 2e-         

so E for the new reaction = [1F(0.77) + 3F(0.04)]/2F = 0.89/2 = 0.445

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry IIT JEE

A decapeptide (Mol.wt 796) on complete hydrolysis gives glycine(Mol.wt 75) , alanine and phenylalanine.  Glycine contributes 47 % to the total weight  of the hydrolysed products . The number of glycine units present in the decapeptide is

Asked By: NITIN RAO
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Joshi sir comment

47 % of (796+162) = 958*47/100 = 450 approx

so no of glycene units = 450/75 = 6 approx

here 162 is taken for the weight of 9 water molecules required for breaking the decapeptides.

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry IIT JEE

A compound contains 3.2% of oxygen . the minimum molecular weight of the compound is ?

Asked By: AMIT DAS
is this question helpfull: 6 2 submit your answer
Joshi sir comment

atomic wt. of oxygen = 16

so m.w./100 = 16/3.2 or m.w. = 500

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