The addition of 3 g of a substance to 100 g CCl4 M - 154 g mol-1 raises the boiling point of CCl4 by 0-60-C- If Kb CCl4 is 5 K mol-1 kg- calculate the molar mass of the substance-Given- Kf CCl4 - 31-8 K kg mol-1 - solution - 1-64 g cm-3-Answer in g-mol

The molar mass of solute is calculated by using the formula M_2=k_b × W 2Δ T_b W_1 Here, subscript 1 refers to solvent and the subscript 2 refers to solute ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Elevation in Boiling Point

The addition ...

Question

The addition of $3 g$ of a substance to $100 g C C l_{4} (M = 154 g m o l^{- 1}$ ) raises the boiling point of $C C l_{4}$ by ${0.60}^{\circ} C$ . If $K_{b} (C C l_{4})$ is $5 K m o l^{- 1} k g$ , calculate the molar mass of the substance.
Given:
$K_{f} (C C l_{4}) = 31.8 K k g m o l^{- 1}$
$ρ (s o l u t i o n) = 1.64 g c m^{- 3}$ .
(Answer in g/mol)

Open in App

Solution

The molar mass of solute is calculated by using the formula $M_{2} = \frac{k_{b} \times W - 2}{Δ T_{b} W_{1}}$
Here, subscript 1 refers to solvent and the subscript 2 refers to solute.
Substitute values in the above expression.
$M_{2} = \frac{5 K m o l^{- 1} k g \times 0.003 k g}{0.6^{o} C \times 0.100 k g} = 0.250 k g / m o l = 250 g / m o l$
Thus, the molar mass of the solute is $250$ g/mol.

Suggest Corrections

Similar questions

Q. The addition of

3 g

of substance to

100 g C C l_{4} (M = 154 g m o l 6 - 1)

raises the boiling point of

C C l_{4}

{0.60}^{\circ} C

K_{b} (C C l_{4})

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

. Calculate:
(a)

the freezing point depression
(b)

the relative lowering of vapour pressure
(c)

the osmotic pressure at

298 K

(d)

the molar mass of the substance
Given

K_{f} (C C l_{4}) = 31.8 k g m o l^{- 1} K

and

ρ (d e n s i t y) o f s o l u t i o n = 1.64 g / c m^{3}

Q. The addition of 3 gm of substance to 100 gm

C C l_{4} (M = 154 g m m o l^{- 1})

raises the boiling point of

C C l_{4}

{0.60}^{\circ} C K_{b}

C C l_{4}

is 5.03 kg

m o l^{- 1}

Calculate :

(a) the freezing point depression

(b) the relative lowering of vapour pressure

(d) the molar mass of the substance

[Given :

K_{f} (C C l_{4}) = 31.8 k g m o l^{- 1} K

and density of solution =

1.64 g m / c m^{3}

]

1.355 g

of a substance dissolved in

55 g

C H_{3} C O O H

produced a depression in the freezing point of

{0.618}^{\circ} C

. Calculate the molar mass of the substance.

(K_{f} = {3.85}^{\circ} C k g m o l^{- 1})

Q. Calculate the mass in grams of an impurity of molar mass

100

{m o l}^{- 1}

which would be required to raise the boiling point of

50

g of chloroform by

0.30

Q. Solution containing 0.73 g camphor (Molar mass

152

g mol

^{- 1})

in 36.8 g of acetone (boiling point

{56.3}^{\circ} C

boil at

{56.55}^{\circ} C

. A solution of 0.564 g unknown compound in same weight of acetone boils at

{56.46}^{\circ} C .

Calculate the molar mass of the unknown compound.

Join BYJU'S Learning Program

The addition of 3g of a substance to 100gCCl4(M=154gmol−1) raises the boiling point of CCl4 by 0.60∘C. If Kb(CCl4) is 5Kmol−1kg, calculate the molar mass of the substance.Given: Kf(CCl4)=31.8Kkgmol−1 ρ(solution)=1.64gcm−3.(Answer in g/mol)

The molar mass of solute is calculated by using the formula M2=kb×W−2ΔTbW1Here, subscript 1 refers to solvent and the subscript 2 refers to solute.Substitute values in the above expression.M2=5Kmol−1kg×0.003kg0.6oC×0.100kg=0.250kg/mol=250g/molThus, the molar mass of the solute is 250 g/mol.