Question

What is the approximate molar mass of a molecular solute if 300 g of the solute in 1000 g of water causes the solution to have a boiling point of 101°C? (Kb = 0.512°C/m; Kf = 1.86°C/m; molar mass of water = 18 g)

Answer 1

If i'm not mistaken, the answer would be 150 amu. 

Answer 2

Answer: The molar mass of solute is 156 g/mol

Step-by-step explanation:

Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.

The equation used to calculate elevation in boiling point follows:

`\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}`

`\Delta T_b` = ? °C

Boiling point of pure water = 100°C

Boiling point of solution = 101°C

Putting values in above equation, we get:

`\Delta T_b=(101-100)^oC=1^oC`

To calculate the elevation in boiling point, we use the equation:

`\Delta T_b=iK_bm`

Or,

`\Delta T_b=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}`

where,

`\Delta T_b` = 1°C

i = Vant hoff factor = 1 (For non-electrolytes)

`K_b` = molal boiling point elevation constant = 0.52°C/m.g

`m_(solute)` = Given mass of solute = 300 g

`M_(solute)` = Molar mass of solute = ?

`W_(solvent)` = Mass of solvent (water) = 1000 g

Putting values in above equation, we get:

`1^oC=1* 0.52^oC/m* (300* 1000)/(M_(solute)* 1000)\\\\M_(solute)=156g/mol`

Hence, the molar mass of solute is 156 g/mol