Calculate the normal boiling point of a sample of sea water found to contain`3.5%` of `NaCl` and `0.13 %`of `MgCl_(2)` by mass. The normal boiling of point of water is `100^(@)C` and `K_(b)("water")= 0.51K kg mol^(-1)` . Assume that both the salts are completely ionised.

Explanation

Mass of NaCl = 3.5 g
No. of moles of NaCl = `3.5/58.5`
Number of ions furnished by one molecule of NaCl is 2.
So, actual number of moles of particles furnished by sodium chloride `=2xx3.5/58.5`
Similarly, actual number of moles of particles furnished by magnesium chloride `=3xx0.13/95`
Total number of moles of particles
`=(2xx3.5/58.5+3xx0.13/95)=0.1238`
Mass of water = (100 – 3.5 – 0.13) = 96.37 g `=96.37/1000kg`
`"Molarity=0.1238/96.37xx1000=1.2846`
`DeltaT_b="Molarity"xxK_b`
`1.2846xx0.51=0.655K`
Hence, boiling point of sea water
= 373.655 K or `100.655^@C`