Question

10gm of a Non-Volatile Solute when dissolved in 100 gm of Benzene raises its boiling point by 1c0. What is the Molecular Mass of the solute? (Kb for Benzene=2.53 K Kg/Mole)​

Answer 1

Answer:

When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by 1ºC then the molecular mass of the solute is (Kb for benzene = 2.53 km⁻¹) 253 g.

Option (d) is correct.

Given,

Kb = 2.53 km^-1

w = 10g

W = 100g

Tb = 1 C

we use the formula,

\begin{gathered}T_b = \dfrac{K_b * w * 1000}{m * W}\\\\m = \dfrac{K_b * w * 1000}{T_b * W}\\\\= \dfrac{2.53 * 10 * 1000}{1 * 100}\\\\=253 g\end{gathered}

T

b

=

m∗W

K

b

∗w∗1000

m=

T

b

∗W

K

b

∗w∗1000

=

1∗100

2.53∗10∗1000

Answer 2

Answer:

253

Explanation:

Given :

w = 10gm of Non - volatile solute

in Benzene = 100 gm

Raise in boiling point = 1°C

Kb of benzene = 2.53 K Kg/Mole

To find :

the Molecular Mass of the solute

Formulae :

$\Delta b = T_b - T_b\, ^\circ = k_b \times m$

m = molality  = $\huge { \text { \frac{\text{weight of the solute}\times 1000}{\text{molecular weight of the solute \times Mass of Solvent in gm}} }}$

Solution :

So from the first formula, let us find the molality

1 = 2.53 × m

=> m =1/2.53.

So,

$\boxed {m = \frac{1}{2.53} }$

From the second formula, let us find the Molecular mass of solute

$m = \frac{1}{2.53} = \frac{10 \times 1000}{M \times 100} \ \ \ [M = \text{Molecular weight of solute}] \\\\=> \frac{1}{2.53} = \frac{100}{M} \\\\=> \underline {M = 253 gm}$

Hence, The Molecular weight of the solute is 253gm

Hope it helps!