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Section:
Chemistry
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© examsnet.com
Question : 11 of 26
Marks:
+1
,
-0
On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm Hg to 640 mm Hg. The depression of freezing point of Benzene (in K) upon addition of the solute is_____
(Given data : Molar mass and the molal freezing point depression constant of benzene are
78
 
g
 
m
o
l
−
1
78\,\mathrm{g\,mol}^{-1}
78
g
mol
−
1
and
5.12
  
K
  
k
g
 
m
o
l
−
1
5.12\;\mathrm{K}\;\mathrm{kg\,mol}^{-1}
5.12
K
kg
mol
−
1
,respectively)
[JEE Adv 2019 P1]
1.02 to 1.03
2.03 to 3.03
1.00 too 1.00
1.25 to 2.50
Validate
Solution:
  
  
10
650
=
  
  
n
B
n
A
+
n
B
\;\;\frac{10}{650} = \;\;\frac{n_{B}}{n_{A}+n_{B}}
650
10
​
=
n
A
​
+
n
B
​
n
B
​
​
n
B
=
10
{}_B=10
B
​
=
10
n
A
=
640
n_{A}=640
n
A
​
=
640
   
m
=
  
  
10
×
1000
640
×
78
m=\;\;\frac{10\times 1000}{640\times 78}
m
=
640
×
78
10
×
1000
​
Δ
t
f
=
  
  
5.12
×
10
×
1000
640
×
78
=
1.025
≃
1.03
\Delta t_{f}=\;\;\frac{5.12\times 10\times 1000}{640\times 78}=1.025\simeq 1.03
Δ
t
f
​
=
640
×
78
5.12
×
10
×
1000
​
=
1.025
≃
1.03
© examsnet.com
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