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.

Options

A. `100.655^(@)C`
B. `99.655^(@)C`
C. `101.655^(@)C`
D. `102.655^(@)C`

Answer

A

Explanation

Mass of `NaCl=3.5g`No. of moles of NACl=`(3.5)/(58.5)`
Numbers of ions furnished by one molecules of `NaCl` is `2`
So, the actual numbers of moles of particals functuion by sodium
chloride =`2xx(3.5)/(58.5)`
Similarly , actual number of mole sof particles furnished by magnesium chloride `=3xx(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=(96.3)/(1000)kg`
Molarity =`(0.1238)/(96.37)xx1000=1.2846`
`DeltaT_(b)` = Molarity `xxK_(b)=1.2846xx0.51=0.655K`