A certain substance X has a normal boiling point of 117.8°C and a molal boiling point elevation constant =Kb·0.63°C·kgmol−1. A solution is prepared by dissolving some urea NH22CO in 900.g of X. This solution boils at 118.3°C. Calculate the mass of urea that was dissolved. Round your answer to 1 significant digit.

42.8 g was the mass of dissolved urea.

Explanation:

Boiling point elevation to solve this:

ΔT = Kb . m

where ΔT → Boiling T° of solution - Boiling T° of pure solvent

Let's replace the data given:

118.3°C - 117.8°C = 0.63°C/m . m

0.5°C / 0.63 m/°C = m → 0.793 mol/kg

Mass of solvent → 900 g → g to kg → 900 g . 1kg/1000 = 0.9 kg

Molality . kg = moles → 0.793 mol/kg . 0.9 kg = 0.714 mol

These are the moles of urea we used. Let's determine the mass dissolved.

0.714 mol . 60 g /1mol = 42.8 g

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.