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New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Molality itself a strength representing terms for solution which does not depend upon the temperature.

 

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Let we take  of solution

Mass of solute = Volume * Density

= 0.5 m l * 1 . 0 5 g m / m l

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

i = 1 . 9

 

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

For A : 100 gm solution 2 gm solute A

M o l a l i t y = 2 / M A 0 . 0 9 8

For B : 100 gm solution 8 gm solute B

M o l a l i t y = 8 / M B 0 . 0 9 2

1 4 . 2 6 1 = M A M B

M B = 4 . 2 6 1 * M A

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

Δ T f = K f * m  

( 1 5 6 1 5 5 . 1 ) = 2 * 1 . 8 M W * 1 0 0 0 ( 6 2 . 5 * 0 . 8 )                

  0.9 =   2 * 1 . 8 * 1 0 0 0 M W * 5 0


M W = 8 0 g m / m o l      

New answer posted

5 months ago

0 Follower 23 Views

V
Vishal Baghel

Contributor-Level 10

P 0 P S P 0 = n s o l u t e n s o l v e n t

P 0 P 0 2 P 0 = n s o l u t e n s o l v e n t

n s o l u t e = n s o l v e n t 2 = 1 0 0 1 8 * 2 = 2 . 7 8 m o l

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

m = w * 1 0 0 0 m o l e c u l a r w t * w s o l v e n t = 1 0 . 2 * 1 0 0 0 1 7 6 * 1 5 0

= 1 0 2 0 0 1 7 6 * 1 5 0 = 0 . 3 8 6

Δ T f = K f * m

3.9 * 0.386

x * 1 0 1 = 1 5 . 0 5 * 1 0 1

Ans =15

New answer posted

6 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

According to Henry's law

PGas=KH*xGas

=0.835=1.67*103* [nCO2nCO2+55.5]

As nCO2<<55.5

nCO2=27.78*103moles

Mass of CO2 gas = nCO2* (MM)CO2

= 1222 * 10-3 gm

New answer posted

6 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

For isotonic solution

? i n j e c t o r = ? B l o o d

? C * 0 . 0 8 2 * 3 0 0 = 7 . 4 7

? C = 0 . 3 ? ? m o l e / l

= 0 . 3 * 1 8 0 g m / l = 5 4 g m / l

New answer posted

6 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

t 1 / 2 = 2 0 0 d a y s = 0 . 6 9 3 k

t = 2 . 3 0 3 k l o g 1 0 N 0 N

83 =  2 . 3 0 3 0 . 6 9 3 * 2 0 0 l o g N 0 N

8 3 = 2 0 0 0 . 3 0 1 0 l o g N 0 N

0.125 = l o g N 0 N

N 0 N = a n t i l o g ( 0 . 1 2 5 )

= 1.333

Activating remaining = N N 0 * 1 0 0

= N 0 1 . 3 3 3 N 0 * 1 0 0

= 75%

 

New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  K = A . e E a / R T = ( 6 . 5 * 1 0 2 ) e 2 6 0 0 0 K / T

E a 8 . 3 1 4 = 2 6 0 0 0

 Ea = 216.164kJ/mol 216

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