Question

calculate the molal elevation constant of water it is being given that 0.2 molal solution of a non volatile non electrolyte increases boiling point of water by 0.104k​

Answer 1

Answer:Density=1.2g/ml=1.2kg/L

The concentration is 1 molar.

1 L of solution contains 1 mole of glucose.

Mass=Density×Volume=1.2kg/L×1L=1.2kg

Thus 1.2 kg of solution contains 1 mole of glucose

mass of solvent = mass of solution - the mass of 1mole glucose

mass of solvent = 1200g−180g

mass of solvent = 1020g=1.02kg.

Molality=  

mass of solvent in kg

number of moles of solute

​  

=  

1.02

1

​  

=0.98

ΔT  

b

​  

=m× K  

b

​  

=0.98 k  

b

​  

≈1 K  

b

​  

 

Explanation:

Answer 2

Answer:

$\bigstar{\bold{K_b=0.52\:K\:kg/mol}}$

Explanation:

$\Large{\underline{\underline{\it{Given:}}}}$

• Molality (m) = 0.2
• Increase in boiling point ($\Delta T_b$)  =0.104 K

$\Large{\underline{\underline{\it{To\:Find:}}}}$

• Molal elevation constant ($K_b$)

$\Large{\underline{\underline{\it{Solution:}}}}$

→ We have to find the molal elevation constant $K_b$ of water

→ $K_b$ is given by the equation,

  $\Delta T_b = K_b\: m$

→ Substituting the given datas we get,

  $0.104=K_b\times 0.2$

→ Simplifying it,

  $K_b=\dfrac{0.104}{0.2}$

 $K_b= 0.52\:K\:kg/ mol$

→ Hence the molal elevation constant of water is 0.52 K kg/mol

$\boxed{\bold{K_b=0.52\:K\:kg/mol}}$

$\Large{\underline{\underline{\it{Notes:}}}}$

→ The increase in the boiling point of a solvent when a compound is added to it is called as the elevation of boiling point.

→ The boiling point of a solution is higher than the pure solvent.

→ $K_b$ is called as the boiling point constant or ebullioscopic constant.

  $\Delta T_b=K_b\:m$