IIT JEE-AIEEE 2012, Entrance Notifications, Study Tips, Engineering Preparation, Practice Test Papers, Experts Advice

Ask iit jee aieee pet cbse icse state board community

Discussion Response Post to: mole concept

Forum Index -> Physical Chemistry -> View Full Question

Author

Message

Titun (1467)

Forum Expert

total posts: 374
Offline

No. of gm atoms of X taken initially = 5 / 60 = 0.083 ( approx. )
No. of gm atoms of Y taken initially = 1.15 x 10 ^23 /(6.023x10^23) = 0.019 (approx)
No. of gm atoms of Z taken initially = 0.03

As given in the problem, 1 atom of X and 2 atoms of Y are required for each 3 atoms of Z.

So, all the gm atoms of Z will be utilized in the reaction but X and Y will be left as residue.

Since, 0.03 gm atoms of Z participates in the reaction, 0.02 moles of Y & 0.01 moles of X are used in the reaction.

Wt. of utilized moles of X in the reaction = 0.01 x 60 = 0.6 gm
Wt. of utilized moles of Z in the reaction = 0.03 x 80 = 2.4 gm

So, the rest of the wt. of the compound i.e (4.4 - 2.4 - 0.6) or 1.4 gm must be due to the 0..02 gm atoms of Y

Wt. of 0.02 gm atoms of Y is 1.4 gm
Hence, wt. of 1 gm atom of Y is 1.4 / 0.02 gm i.e 70 gm
Therefore, 1 atom of Y weighs 70/(6.023x10^23) gm = 70 amu

Therefore, atomic weight of Y is 70 amu

Cheers !!

You never know what is enough till you know what is more than enough. Titun

Like 0 people liked this

Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info