N-factor calculation practice!

Chemistry Level 5

Calculate the n-factor of all the underlined species and Find the sum of all n-factors.

Your answer can be represented as $\frac{M}{N}$ , where $M$ and $N$ are coprime positive integers.

Enter your answer as $M + N$ .

Original

These reactions are some of the important reaction of p-block elements, it is suggested (by me) that you should also learn these reactions for JEE.

Here is the vast collection of questions of many different varieties that I have posted uptill now . You will enjoy solving them , Have a look Click Here :)

The answer is 1933.

1 solution

Aniket Sanghi
Jan 30, 2017

Remember the Basic Defination - n-factor is the total no. of electrons gained or loosed $per$ molecule of the sprcies .

Answers are

a - 2/3
b - 6
c - 5
d - 7
e - 3
f - 4
g - 3/4
h - 4/5
i - 3
j - 1.

@Aniket Sanghi

Can you please elaborate part c and d ( can't we say that equivalent mass of a substance is the mass which reacts to produce 8gm of oxygen and then find n factor) Also in last part , isn't n factor in non redox change calculated by finding no. Of ions exchanged per molecule? Or we can say that since h2so4 replaces 1 H+ hence its n factor is 1 and since no of equivalents of h2so4 equals no. Of equivalents of nano3 , n factor of nano3 is also 1 Pls correct if I m wrong because this is how my sir taught it in fiitjee.

Hargun Preet Singh - 4 years, 4 months ago

I was also confused uptill PT3 about n-factor but in part test 3 one question is there regarding Def of n-factor , check it out ! That cleared aLloyd my doubts !

The basic method is

Calc the total no. Of electrons gained (or loosed) in a met reaction , then divide it by the no. Of moles of the species (n-factor is asked for ) , no. Of moles that appear in balanced equation .

The way you are dealing is the way suitable for simple oxidation and reduction . The questions appearing above are in general disproportion.

For c , I is reducing and others O and N are oxidising . So if you consider oxidation changes you have to consider two mol, so better will be consider reduc change . I is changing by 5. In one mol we have 1 I. Hence we have 5 electron gained per mol of our molecule . Hence n-factor 5! Similarly for d.

Aniket Sanghi - 4 years, 4 months ago

@Aniket Sanghi Sorry to ask here but can u share ur score in fiitjee open test held recently

Hargun Preet Singh - 4 years, 3 months ago

My got bad this time it's only 185/372 this time :(

Aniket Sanghi - 4 years, 3 months ago

These reactions are some of the very important and basic reactions of P-Block one should memorize. But it doesnt contain reactions of other groups except 15.

Md Zuhair - 2 years, 8 months ago

Please explain part b

Pavani Bindal - 1 year, 11 months ago

explain as in you need solution?

Md Zuhair - 1 year, 1 month ago

hi aniket could you please give a detailed solution of part (f) as i am getting the answer as 2 rather than 4 and i cant figure out my mistake

Aryan katiyar - 1 year, 1 month ago

N-factor calculation practice!

These reactions are some of the important reaction of p-block elements, it is suggested (by me) that you should also learn these reactions for JEE.

Here is the vast collection of questions of many different varieties that I have posted uptill now . You will enjoy solving them , Have a look Click Here :)

The answer is 1933.

1 solution

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