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some doubts from chem --- Anubhav

some doubts from chem --- Anubhav

by Anubhav Sharma -
Number of replies: 1

1. The n-factor of CuCN in the given reaction is

    CuCN -------------> Cu2+   +   CO2   +   HNO3

    (A) 11

    (B) 7

    (C) 9

    (D) 3

Answer is (A) . Explain it please.

2. Which of the following is isoelectronic

    (|) CH3+   (II) H3O+  (III) NH3  (IV)  CH3-

    (A) I and II

    (B) III and IV

    (C) I and III

    (D) II , III and IV

Answer is (D) . Kindly explain in brief. What is isoelectronic and how to solve such questions ?

3. An element X forms an oxide with 23.12 % oxygen and a chloride with 43.25 % chlorine. The ratio of valencies exhibited by X in its oxide and chloride is

    (A) 7:3

    (B) 5:3

    (C) 5:4

    (D) 7:4

Answer is (A). Explain in brief.

4. Al2 (SO4)3 . x H2O has 8.1 % of aluminium by mass . The value of x is

    (A) 6

    (B) 9

    (C) 12

    (D) 18

Answer is (D). Kindly explain in brief.

Thanks a Lot !

Your Good Friend

Anubhav

In reply to Anubhav Sharma

Re: some doubts from chem --- Anubhav

by Suvarthi Datta -
(1.) Here Cu+ -> Cu++ (change in oxidation numnber is 1)
Change in oxidation number of N is 10 (from -5 to +5)
C remains as +4 . There is no change in oxidation number of C

Therefore , n-factor=(1) + [5-(-5)] + 0
= 11 .

n-factor is the total change in oxidation number of 1 nole of a substance .


(2.) Isoelectronic means " SAME NUMBER OF ELECTRONS " .

CH3+ has (6+3-1)= 8 electrons

H3O+ has (3+8-1) = 10 electrons

NH3 has (7+3)= 10 electrons

CH3- has (6+3+1) = 10 electrons

So H3O+ , NH3 , CH3- are isoelectronic .


(3.) Let the formula of the oxide be M2Ox and the chloride be
M(Cl)y . Let atomic mass of M be m .
Molecular mass of the oxide = (2m+16x)
Molecular mass of the chloride = (m+35.5y)

Now , (23.12/100)=(16x)/(2m+16x)
=> m = 26.6x ------------(1)

Again , (43.25/100)=(35.5y)/(m+35.5y)
=> m = 46.58y------------(2)

From (1) and (2) :

=> x/y = 1.75
=> x : y = 7 : 4

My answer is (D) . Please let me know if I am wrong .

(4.) Molecular mass of the Al2(SO4)3.xH2O = (18x+342)

Now , (8.1/100)*(18x+342) = 54

=> x = 18