- Organic Chemistry
- Aldehydes and Ketones
- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
- Aromatic Chemistry
- Carbohydrates, Amino acids, protein, Vitamin and Fat
- Carboxylic acids and its derivatives
- Chemistry in daily life
- General Mechanism in organic compounds
- Hydrocarbons
- Nomenclature and isomerism

# 18 - Mole Concept and Stoicheometry Questions Answers

A sample of fuming HNO3 is labelled as 110% . assume that the labelling of fuming is similar the labelling of oleum . What is the % of free N2O5 in the sample??

plz annex explanatn also ans is 60%

**Asked By: SARIKA**

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**Joshi sir comment**

x % HNO_{3 }--------------------- 100 + (18/108)*x % of fuming HNO_{3}

so 18x/108 = 10

or x = 60 %

**The density of a gas is 3.80 g/l at STP. Calculate its density at 27°C and 700 torr pressure.**

**Asked By: SWATI KAPOOR**

**submit your answer**

**Joshi sir comment**

P = dRT/M

for first condition

760 = 3.8R273/M

and 700 = dR300/M

divide and get the answer

Two oxides of a metal M contain respectively 22.53% and 30.38% of O_{2}. If the formula of first oxide is MO, the formula of second oxide is

(a) M_{2}O_{3} (b) MO_{3} (c) M_{2}O (d) M_{2}O_{4}

**Asked By: SWATI KAPOOR**

**submit your answer**

**Joshi sir comment**

M O

77.47 22.53

m w M 16

formula for first oxide is MO

so 77.47/M = 22.53/16

so M = 77.47*16/22.53 = 55

M O

69.62 30.38

m w 55 16

so M:O = 69.62/55:30.38/16 = 1.27:1.90

1.9/1.27 = 1.5

so formula will be M_{2}O_{3}

24.5g KClO_{3} heated to give O_{2}, which is allowed to react completely with H_{2} to form H_{2}O. This H2 comes from Zn and H_{2}SO_{4} reaction. The amount of Zn required for the purpose is

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

2KClO_{3} ----------------------> 2KCl + 3O_{2}

3O_{2} + 6H_{2} --------------------> 6H_{2}O

6Zn + 6H_{2}SO_{4} -------------------> 6ZnSO_{4} + 6H_{2}

so 2 mole _{ }KClO_{3 }----------------------------- 6 mole Zn

now solve

If the potential energy of hydrogen electron is -3.02eV in which of the following energy levels is electron present =

1^{st} , 2^{nd} , 3rd or 4^{th }

**Asked By:**

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**Joshi sir comment**

Energy = -13.6/n^{2} , Potential Energy = -27.2/n^{2}

for n = 3 we get -3.02

in this answer infinite is zero potential energy level

A 0.5 g sample of sodium nitrate on heating gives 44.8ml of O_{2} at STP. % purity of the sample is.............

my answer is 64% but the correct answer is 68%. HOW???

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

NaNO_{3 }−−−−> NaNO_{2} + 1/2 O_{2}

molar mass of sodium nitrate = 85

so 85 gm will give 11200 ml O_{2}

so x gm will give 11200*x/85 = 2240x/17

on comparing with 44.8

we get 2240x/17 = 44.8

so x = 44.8*17/2240 = 17/50 = 34/100 = 0.34 gm

means 0.34 gm sodium nitrate is pure in 0.5 gm. sample so % purity = 0.34*100/0.50 = 68%

Equivalent weight of A_{2}O in the given reaction is A_{2}O = AO (molar mass of A_{2}O = M)

**Asked By: SWATI KAPOOR**

**submit your answer**

**Joshi sir comment**

O.N. change = 1-2 = -1

2 A are present so net change = 2

so e.w. = M/2

Mole fraction of C_{2}H_{5}OH in aqueous solution is 0.25. If the density of solution is 0.967g/ml. calculate molarity and molality.

**Asked By: NEHA RANI**

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**Joshi sir comment**

0.75 mole water contains ...................................... 0.25 mole C_{2}H_{5}OH

so 0.75*18 gm contains ....................................... 0.25 mole

now calculate moles of alcohol in 1000 gm water. It will be molality

similarly

0.75 mole water contains ...................................... 0.25 mole C_{2}H_{5}OH

so 0.75*18 gm contains ....................................... 0.25*46 gm

so 0.75*18+0.25*46 gm solution contains ........................... 0.25 mole alcohol

so (0.75*18+0.25*46)*0.967 ml solution contains ............................. 0.25 mole alcohol

now calculate moles of alcohol in 1000 ml solution. It will be molarity

Sir the formula for converting volume in STP condition is given in my book if volume is not given in STP condition. But my problem is that I don't know how to use it . please explain me this formula and I request you to give me Ques. related to this formula.

**Asked By: HEMA RANI**

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**Joshi sir comment**

Formula for converting volume at temperature T to STP is

P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}

Two oxides of a metal contain 72.4% and 70% of metal respectively. If formula of 2nd oxide is M_{2}O_{3}, find that of the first?

Sir the answer is M_{3}O_{4}. But I don't know the steps for doing this type of question, so pls explain me each step!!!

**Asked By: HEMA RANI**

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**Joshi sir comment**

for second oxide ratio of metal and oxygen = 70:30

let molcular mass of metal = M so 70/30 = 2M/3*16

or 7/3 = 2M/48 so M = 24*7/3 = 56

now for first oxide ratio of metal and oxygen = 72.4/27.6 (By weight)

so ratio by no. of atoms = (72.4/56)/(27.6/16) = 72.4*16/27.6*56 = 1158.4/1545.6 = 3/4