Question

Boiling point of water at 750 mmHg is 99.63⁰C. How much sucrose is to be added to 500 g of water such that it boils at 100⁰C. Kb for water 0.52 K Kg mol-1

Answer 1

Answer:

121.67 g is to be added to 500 g of water

Explanation:

Given that:

Pressure = 750 mmHg

Temperature T₁= 99.63⁰C = (273 + 99.63 ) = 372.63K

mass of water = 500 g

Temperature T₂  = 100⁰C = ( 273 + 100) K = 373 K

where;

Kb for water 0.52 K Kg mol-1

For sucrose; C₁₂ H₂₂ O₁₁

Molar mass =  ( 12 × 12 )+ ( 1 × 22 ) + ( 16 × 11 )

Molar mass = 342 g/mol

ΔT = T₂  - T₁

ΔT = (373 - 372.63)K

ΔT = 0.37 K

∴ the amount of sucrose to be added to  500 g of water is:

 $= \dfrac{0.37\times 342 \times 500}{0.52 \times 1000 }$

$= \dfrac{6327}{52}$

= 121.67 g

Thus; 121.67 g is to be added to 500 g of water