Question

Calculate the boiling point of a solution containing 0.456 gram of camphor

Answer 1

Answer:

i think the question is incomplete

if it is the question is

Calculate the boiling point of a solution containing 0.456g of camphor mol.mass 152,dissolved in 31.4g of acetone boiling point=56.30ÂC,if the molecular elevation constant per 100g of acetone is 17.2Â.

Explanation:

Dear Student,

As from the given data:

Figure moles of camphor  

Molar mass = 152 g/mol  

No. Of mole = 0.456/152 = x  

Mass of CH3)2CO = 31.4 g  

Molality = x*1000/31.4 =m  

Rise of BP = k*m  

K= 17.2 for each 100g  

= 1.72 Per Kg

Rise = 1.72 * x*1000/31.4

BP = 56.46⁰ C