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New answer posted•a year ago
A
a year ago

C 6 H 1 2 O 6 → G l u c o s e

M a s s     o f     C m a s s     o f     g l u c o s e = 7 2 1 8 0

% of C = 10.8 = M a s s     o f     C m a s s     o f     s o l u t i o n × 1 0 0

1 0 . 8 × 2 5 0 1 0 0 m a s s     o f     C

∴ M o l a r i t y = 0 . 3 7 5 0 . 1 8 2 5 = 2 . 0 5 5 = 2 . 0 6

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New answer posted•a year ago
V
a year ago

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

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New answer posted•a year ago
New question posted•a year ago
New answer posted•a year ago
V
a year ago

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

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New answer posted•a year ago
V
a year ago

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

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New answer posted•a year ago
New answer posted•a year ago
P
a year ago

Volume of H2 adsorbed = nRTP=2×0.083×3002×1=24.9  lit=24900  ml

Therefore volume of gas adsorbed per gram of the adsorbent = 249002.5=9960

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New answer posted•a year ago
New answer posted•a year ago
V
a year ago

Mass = d × v = 1.02 × 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

Δ T f = i × K f × m

i = 1 + a for acetic acid

0.0198  = (1 + a) × 1.85 × 0.0102

α = 0.04928 = 5%

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