- Organic Chemistry
- Aldehydes and Ketones
- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
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- Carbohydrates, Amino acids, protein, Vitamin and Fat
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- Chemistry in daily life
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# 13 - Atoms and molecules Questions Answers

At 752 mm Hg of total Pressure and 301 K Temperature 150 ml H_{2} is produced and collected over the water, when current is passed through acidic water for 1 hour. How many Amp. current should have passed? ( at 301 K temperature, The pressure of water is 28 mm Hg )

A) 0.320 A

B) 310 A

C) 321 A

D) 0.305 A

please mention correct answer.

**Asked By: BONEY HAVELIWALA**

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**Solution by Joshi sir**

by formula PV/T = constant

(752-28)*150/301 = 760*V/273

solve for V, this will be the equivalent volume at STP

now convert this volume in gm by the formula 2V/22400

then finally m = zit, where

z = 1/96500

solve it for i

if there are Three Fe^{+2 } For a Fe^{+3} in Fe_{x}O, What is the value of x?

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

3FeO.1Fe_{2}O_{3} = Fe_{5}O_{6} = Fe_{5/6}O

the first line in the balmer series in the hydrogen spectrum will have the frequency

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

f/c = R[ 1/2^{2} - 1/3^{2} ]

R and f are Rydberg constant and frequency

if radius of first bohr orbit of hydrogen atom is X, then de Broglie wavelength of electron in 4 th orbit is

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

radius of fourth orbit = 16X

now put 2πr = 4λ

**Asked By: SACHIN TRIPATHI**

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**Joshi sir comment**

Consider about the amount of energy required. It can gain from its surrounding

A hydrogen like system has ionisation enthalpy 11808KJ/mol. find the no of protons in the nucleus of the system?

**Asked By: MOHIT SAINI**

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**Joshi sir comment**

A hydrogen like system has ionisation enthalpy 11808KJ/mol

on converting it into eV we get [11808*1000]/[6.023*10^{23}*1.6*10^{-19}] = 122.53 eV

now formula for this energy is [-Z^{2}/n^{2}]13.6 eV

for n = 1, Z will be 3

the position of both an electron and a helium atom is known within 1nm. further the momentum of the electron is known within 5*10^{-26} kgm/s. the minimum uncertainity in the measurement of the momentum of the helium atom is

**Asked By: SHIVANI MALIK**

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**Joshi sir comment**

according to Heisenberg principle product of uncertainity in position and momentum of a particle is greater than or equal to h/4π, since uncertanity in position of both the particle are same so same will be the momentum uncertainty minima.

the frequency of radiation emitted when the electron falls n=4 to n=1 in a hydrogen atom will be (given ionisation energy of H =2.18*I0^{-18} J/atom )

ans=3.08*10^{15 }

**Asked By: SHIVANI MALIK**

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**Joshi sir comment**

for n = 1 E = 2.18*10^{-18}

for n = 4 E = 2.18*10^{-18}/16

so hν = 15*2.18*10^{-18}/16

so ν = 15*2.18*10^{-18}/16*6.634*10^{-34}

so ν = 3.08*10^{15}

A mixture of FeO and Fe304 when heated in air to constant weight gains 5% in its weight. The % of FeO in the sample

**Asked By: AMIT DAS**

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**Joshi sir comment**

let FeO = x gm and FeO+Fe_{2}O_{3}= y gm so FeO = [72/(72+160)]*y and Fe_{2}O_{3 }= [160/(160+72)]*y

2FeO + 1/2O_{2 ----------> }Fe_{2}O_{3}

so amount of Fe_{2}O_{3 }formed by x gm FeO = 160x/144

similarly amount of Fe_{2}O_{3 }formed by [72/(72+160)]*y gm FeO = 160 [72/(72+160)]*y/144

so total Fe_{2}O_{3} = 160x/144 + 160 [72/(72+160)]*y/144 + [160/(160+72)]*y = 160x/144 + [160*72/232]y/144 + 160y/232

now according to the given condition {[160x/144 + [160*72/232]y/144 + 160y/232}/{x+y) = 105/100

now solve

what is the relation b/w E_{1} &E_{2} whose respective wavelength r 400& 800 A^{o} . SIR I WANT TO KNOW THE RELATn b/w ENERGY & WAVELENGTH . ARE THEY DIRECTLY PROPTIONAL ? ANS OF THIS QUES IS E_{1}=2E_{2}

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

E = hc/λ

so E is inversely proportional to λ