217 - Chemistry Questions Answers

1.0 ml of dilute solution of NaOH is added to 100 ml of a buffer of pH 4. The pH of the resulting solution 

(a) becomes 7.0          

(b) becomes 9.0

(c) becomes 3.0

(d) Remains practically unchanged

Asked By: SWATI KAPOOR
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Joshi sir comment

it is strong acidic buffer so answer is (d)

10 ml of M/200 H2SO4 is mixed with 40 ml of M/200 H2SO4. The ph of the resulting solution is

Asked By: SWATI KAPOOR
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Joshi sir comment

M of H2SO4 = 1/200

so [H+] = 2*1/200 = 1/100 = 10-2

now calculate pH

 

 

 

The pH of solution at 25°C which has twice as many hydroxide ion as in pure water at 25°C , will be

Asked By: SWATI KAPOOR
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Joshi sir comment

no of hydroxide ion in pure water = 10-7

so according to the given condition no. of hydroxide ions in solution = 2*10-7

so pOH = -log2*10-7

so pH = 14 - pOH 

The ph of 0.1 M HCN is 5.2. What is the Ka?

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Joshi sir comment

HCN ------------------->  H+ + CN-

0.1                                 0          0

0.1-x                              x           x

pH = 5.2

so -log[H+] = 5.2 

solve for [H+], it will give x

then calculate Ka

 

ΔHf  C2H4 = 12.5 kcal

Heat of atomization of C = 171 kcal  

Heat of atomization of H = 52.1 kcal  

Bond energy CH = 99.3 kcal .

What is C=C bond energy?

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Joshi sir comment

2 heat of atomization of C + 4 heat of atomization of H - 4 bond energy of C-H - 1 bond energy of C=C = heat of formation of C2H4

2*171 + 4*52.1 - 4*99.3 - x = 12.5

solve

How Is the value of n calculated for nerst equation??

 

Asked By: AAYUSH TANEJA
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Joshi sir comment

n is the no. of electrons in anode or cathode reaction.

If no. of electrons are different then make these no. common

for ex. if reactions are 

             Al ---------------> Al+++ + 3e-

and     Cu++ + 2e- ------------> Cu

then multiply (1) to 2 and (2) to 3

after that electrons will be 6 in both reactions so n = 6

which of the following has the maximum value for van der waals constant b - H2O (g) , AlCl3 (g), Cl2 (g)

Asked By: SWATI KAPOOR
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Joshi sir comment

b depends on size so on the basis of size order will be AlCl3>Cl2>H2O

This answer is given by Sarika and is correct answer

 

the type of interaction present between NO3- ion and I2 molecule is ion-dipole interactions or ion-induced dipole interactions.Explains the difference between the two also.

 

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Joshi sir comment

Iis non polar molecule so it will be ion-induced dipole interactions 

Arrange the following molecules in the increasing order of ionic characters,

LiF, K2O, N2, SO2, ClF

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Joshi sir comment

N2SO2ClF3, K2O, LiF

Solution by Joshi sir

N2SO2ClF3, K2O, LiF

This order is due to difference of electronegativity of atoms in a molecule

more difference more will be the ionic character

The ratio of positive charge on H- atom in HX to that in HY from the following data is 

Dipole moment of HX = 1.14 D and HY = 0.38 D

Bond length of HX = 127pm and HY = 170pm

Options are - (i) 3:4 ,   (ii) 4:1  ,  (iii) 1:4

Asked By: SWATI KAPOOR
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Joshi sir comment

dipole moment = charge * bond length

now solve

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