Question

What mass in grams of a molecular substance (molar mass = 50.0 g/mol) must be added to 500 g of water to produce a solution that boils at 101.54 oC? (Kbp = 0.512 oC/m for water.)

Answer 1

Answer: The mass of the substance that must be added is 75.2 grams.

Step-by-step explanation:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing 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 = 101.54°C

Putting values in above equation, we get:

`\Delta T_b=(101.54-100)^oC=1.54^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` = 1.54°C

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

`K_b` = molal boiling point elevation constant = 0.512°C/m.g

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

`M_(solute)` = Molar mass of solute = 50.0 g/mol

`W_(solvent)` = Mass of solvent (water) = 500 g

Putting values in above equation, we get:

`1.54^oC=1* 0.512^oC/m* (m_(solute)* 1000)/(50.0g/mol* 500)\\\\m_(solute)=(1.54* 50.0* 500)/(1* 0.512* 1000)=75.2g`

Hence, the mass of substance that must be added is 75.2 grams.