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Byju's Answer
Standard IX
Chemistry
Molecules and Atomicity
Calculate the...
Question
Calculate the mass (in grams) of one molecule of
C
H
3
C
O
O
H
.
(Molar mass of
C
H
3
C
O
O
H
=
60.06
g
/
m
o
l
)
A
9.97
×
10
−
23
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B
60.06
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C
9.97
×
10
23
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D
None of above
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Solution
The correct option is
D
9.97
×
10
−
23
The molar mass of
C
H
3
C
O
O
H
is given as
60.06
g
/
m
o
l
We need to find the mass of
1
molecule of
C
H
3
C
O
O
H
We can use this formula:
N
u
m
b
e
r
o
f
a
t
o
m
s
=
m
a
s
s
m
o
l
a
r
m
a
s
s
×
A
v
o
g
a
d
r
o
n
u
m
b
e
r
We can substitute the following values in the formula:
N
u
m
b
e
r
o
f
a
t
o
m
s
=
1
M
o
l
a
r
m
a
s
s
=
60.06
g
/
m
o
l
A
v
o
g
a
d
r
o
n
o
.
=
6.022
×
10
23
N
u
m
b
e
r
o
f
a
t
o
m
s
=
m
a
s
s
m
o
l
a
r
m
a
s
s
×
A
v
o
g
a
d
r
o
n
u
m
b
e
r
∴
1
=
m
a
s
s
60.06
×
6.022
×
10
23
m
a
s
s
=
60.06
6.022
×
10
23
=
9.973
×
10
−
23
g
Suggest Corrections
0
Similar questions
Q.
One gram of charcoal adsorbs
C
H
3
C
O
O
H
from
100
mL
of
0.6
M
C
H
3
C
O
O
H
aqueous solution to form a monolayer, and thereby the molarity of
C
H
3
C
O
O
H
reduces to
0.58
. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal per gram
=
3.0
×
10
2
m
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. Take
N
A
=
6
×
10
23
Q.
(a) What is meant by the 'molar mass' of a substance ? State the unit in which molar mass is usually expressed.
(b) Calculate the molar masses of the following substances. Write the results with proper units.
(i) Ozone molecule, O
3
(ii) Ethanoic acid, CH
3
COOH
Q.
1.355
g
of a substance dissolved in
55
g
of
C
H
3
C
O
O
H
produced a depression in the freezing point of
0.618
∘
C
. Calculate the molar mass of the substance.
(
K
f
=
3.85
∘
C
k
g
m
o
l
−
1
)
Q.
One gram of charcoal absorbs 100 ml 0.5 M
C
H
3
C
O
O
H
to form a mono layer, and thereby the molarity of
C
H
3
C
O
O
H
reduces to 0.49. Calculate the surface area of the charcoal absorbed by each molecule of acetic acid. Surface area of charcoal =
3.01
×
10
2
m
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/
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Q.
Calculate
[
H
+
]
from a
C
H
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C
O
O
H
in a solution which is
10
−
1
M
in
H
C
l
and
10
−
3
M
in
C
H
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C
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.
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