Question

calculate the vapour pressure of solution having 3.42 gm cane sugar in 180 gm water at 40°C and 100°C. Given that boiling point of water 100°C and heat of vapourisation is 10KCal/mole in the given temperature range​

Answer 1

Answer:

58 mm

Explanation:

At 100

o

C, Vapour pressure of pure water (P

0

)=760mm

∵

P

S

P

0

−P

S

=

m×W

w×M

.....(i)

∴

P

S

760−P

S

=

342×180

3.42×18

∴P

S

=759.2mm

Also we have, 2.303log

P

1

P

2

=

RT

1

T

2

ΔH[T

2

−T

1

]

∵P

2

=760mm;T

2

=373K;

T

1

=313K and ΔH=10kcalmol

−1

∴ From eq. (ii), 2.303log

P

1

760

=

2×10

−3

10

×

373×313

[373−313]

∴P

1

=58.2mm at 313 K

Now at 40

o

C:

P

S

P

0

−P

S

=

m×W

m×W

P

S

58.2−P

S

=

352×180

3.42×18

∴P

S

=58.14mm

For 0.2 molal solution: P

H

2

O

0

=58.2mm at 40

o

C

∵w/m=0.2 and W=1000g;M=18

∴

P

S

58.2−P

S

=

1000

0.2×18

∴P

S

=57.99mm≈58mm

Answer 2

${ \green{ \underline{ \huge{ \mathfrak \red{♡Answer}}}}}$

Plzzz refr the attachment fr ur answer...

plzz mention if it's wrong...

${ \red{ \fbox{ \green{ \blue{ \underline{ \mathcal \purple{hope \: this \: helps \: u}}}}}}}$

☺✌

mate I'm really sry to say but ur answer fr my last question is absolutely wrong...

plzz refr the ans by another mate in my same question hope u understand ur mistake...

but tq fr ur effort... ☺