A non-volatile solute- -A- tetramerises in water to the extent of 80 - -2 -5 g of -A- in 100 g of water- lower the freezing point by 0-3- C- The molar mass of A in g - mol is- K f - for water -1-86 Kkgmol -1A- 221B- 62C- 354D- 155

The correct option is B 62Δ T_f = i× K_f× m, α = 0.8 4A⇌ A_4 i = 4 4α+α/4 Van 't Hoff factor (i) = 0.4 Given that Δ T = 0.3 K, m = 25/M (M = Molar mass of A) ...

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Standard XII

Chemistry

Elevation in Boiling Point

A non-volatil...

Question

A non-volatile solute, 'A' tetramerises in water to the extent of 80%. 2.5g of 'A' in 100 g of water, lower the freezing point by ${0.3}^{o} C .$ The molar mass of $A$ in $(g / m o l)$ is: $(K_{f}$ for water = 1.86 $K k g m o l^{- 1})$

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Solution

The correct option is A 62
$Δ T_{f} = i \times K_{f} \times m$ , $α = 0.8$
$4 A ⇌ A_{4}$

$i = \frac{4 - 4 α + α}{4}$
Van 't Hoff factor (i) = 0.4
Given that $Δ T = 0.3 K, m = \frac{25}{M}$
(M = Molar mass of A),
$0.3 = 0.4 \times 1.86 \times \frac{25}{M} M = 62 g / m o l$

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Similar questions

Q. Substance

A

tetramerises in water to the extent of 80%. A solution of

2.5 g

A

100 g

of water lowers the freezing point by

{0.3}^{\circ} C

. The molar mass of

A

is:

(

K_{f}

for water

=

1.86 K

k g

m o l^{- 1}

)

Q. Substance

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tetramerises in water to the extent of

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100 g

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{0.3}^{o} C

. The molar mass of

A

is (

K_{f}

for water

= 1.86 K K g m o l^{- 1}

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Q. A certain substance 'A' tetramerises in water to the extent of

80 %

. A solution of

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g of

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100

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{0.3}^{o} C

. The molar mass of A is:

Q. A certain substance 'A' tetramerises in water to the extent of 80%. A solution of

2.5 g

of A in

100 g

of water lowers the freezing point by

{0.3}^{\circ} C .

The molar mass of A in (g/mol) is:
(Molal depression constant of A is

1.86 K k g m o l^{- 1}

)

The freezing point of a 0.05 molal solution of a non electrolyte in water is (K_f = 1.86 k kg mol^-1)

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A non-volatile solute, 'A' tetramerises in water to the extent of 80%. 2.5g of 'A' in 100 g of water, lower the freezing point by 0.3oC. The molar mass of A in (g/mol) is: (Kf for water = 1.86 Kkgmol−1)

The correct option is A 62ΔTf=i×Kf×m, α=0.8 4A⇌A4 i=4−4α+α4 Van 't Hoff factor (i) = 0.4 Given that ΔT=0.3K, m=25M (M = Molar mass of A), 0.3=0.4×1.86×25MM=62 g/mol

A non-volatile solute, 'A' tetramerises in water to the extent of 80%. 2.5g of 'A' in 100 g of water, lower the freezing point by ${0.3}^{o} C .$ The molar mass of $A$ in $(g / m o l)$ is: $(K_{f}$ for water = 1.86 $K k g m o l^{- 1})$

The correct option is A 62
$Δ T_{f} = i \times K_{f} \times m$ , $α = 0.8$
$4 A ⇌ A_{4}$

$i = \frac{4 - 4 α + α}{4}$
Van 't Hoff factor (i) = 0.4
Given that $Δ T = 0.3 K, m = \frac{25}{M}$
(M = Molar mass of A),
$0.3 = 0.4 \times 1.86 \times \frac{25}{M} M = 62 g / m o l$