You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Lewis Kossel Structure

Calculate the...

Question

Calculate the mass of ascorbic acid $(C_{6} H_{8} O_{6})$ to be dissolved in $75$ g of acetic acid to lower its melting point by ${1.5}^{0} C$ . (Kf for acetic acid is $3.9$ K kg ${m o l}^{- 1}$ ).

Open in App

Solution

$w_{2} = \frac{M_{2} w_{1} Δ T_{f}}{1000 K_{f}}$
$= \frac{176 \times 75 \times 1.5}{1000 \times 3.9}$
$= 5.08 g$

Suggest Corrections

Similar questions

Q. Calculate the mass of ascorbic acid (Vitamin C.

C_{6} H_{8} O_{6}

) to be dissolved in 75 g of acetic acid to lower its melting point by

1 {.5}^{\circ} C . K_{f} = 3.9 K kg mo l^{- 1}

Q. Calculate the mass of ascorbic acid

(V i t a m i n C, C_{6} H_{8} O_{6})

to be disssolved in 75 g of acetic acid to lower its melting point by

1.5. K_{f} = 3.9 K^{- 1}

Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in
75 g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol⁻¹.

Q. The freezing point of a solution containing

0.2

g of acetic acid in

20.0

g benzene is lowered by

{0.45}^{o} C

. The degree of association of acetic acid in benzene is:

[Assume acetic acid dimerizes in benzene and

K_{f}

for benzene

= 5.12

K kg mol

^{- 1}

]

Q. The freezing point of benzene decreases by

{0.45}^{0} C

when

0.2

g of acetic acid is added to

20

g of benzene. If acetic acid is associates to form a dimer in benzene, percentage association of acetic acid in benzene will be:
(

K_{f}

for benzene = 5.12 K kg

m o l^{- 1}

)

Join BYJU'S Learning Program

Calculate the mass of ascorbic acid (C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.50C. (Kf for acetic acid is 3.9 K kg mol−1).

w2=M2w1ΔTf1000Kf=176×75×1.51000×3.9=5.08g

Calculate the mass of ascorbic acid $(C_{6} H_{8} O_{6})$ to be dissolved in $75$ g of acetic acid to lower its melting point by ${1.5}^{0} C$ . (Kf for acetic acid is $3.9$ K kg ${m o l}^{- 1}$ ).

$w_{2} = \frac{M_{2} w_{1} Δ T_{f}}{1000 K_{f}}$
$= \frac{176 \times 75 \times 1.5}{1000 \times 3.9}$
$= 5.08 g$