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 = 3.5g
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 `= 2 xx (3.5)/(58.5)`
Similarly, actual number of moles of particles furnished by magnesium chloride `= 3 xx (0.13)/(95)`
Total number of moles of particles `= (2xx(3.5)/(58.5)+3xx(0.13)/(95))`
`=0.1238`
Mass of water `=(100-3.5-0.13)=96.37g = (96.37)/(1000)kg`
Molality `= (0.1238)/(96.37)xx1000=1.2846`
`DeltaT_(b)=` Molaity `xx K_(b)`
`=1.2846xx0.51=0.655K`
Hence, boiling point of sea water `= 373.655 K or 100.655^(@)C`.