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--------------------- 100 + (18/108)*x % of fuming HNO3

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
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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 O2. If the formula of first oxide is MO, the formula of second oxide is

(a) M2O3         (b) MO3         (c) M2O         (d) M2O4

Asked By: SWATI KAPOOR
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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 M2O3

24.5g KClO3 heated to give O2, which is allowed to react completely with H2 to form H2O. This H2 comes from Zn and H2SO4 reaction. The amount of Zn required for the purpose is

Asked By: SWATI KAPOOR
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Joshi sir comment

2KClO3 ----------------------> 2KCl + 3O2

3O2 + 6H2 --------------------> 6H2O

6Zn + 6H2SO4 -------------------> 6ZnSO4 + 6H2

so 2 mole  KClO3  -----------------------------  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 =

1st , 2nd , 3rd  or  4th 

Asked By:
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Joshi sir comment

Energy = -13.6/n2 ,  Potential Energy = -27.2/n2

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 O2 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

NaNO3  −−−−>    NaNO2 + 1/2 O2

molar mass of sodium nitrate = 85

so 85 gm will give 11200 ml O2

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 A2O in the given reaction is A2O = AO             (molar mass of A2O = M)

Asked By: SWATI KAPOOR
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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 C2H5OH in aqueous solution is 0.25. If the density of solution is 0.967g/ml. calculate molarity and molality.

Asked By: NEHA RANI
is this question helpfull: 6 5 read solutions ( 2 ) | submit your answer
Joshi sir comment

0.75 mole water contains ...................................... 0.25 mole C2H5OH

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 C2H5OH

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

P1V1/T1 = P2V2/T2

Two oxides of a metal contain 72.4% and 70% of metal respectively. If formula of 2nd oxide is M2O3, find that of the first?    

Sir the   answer is M3O4. 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

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