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Calculate the normal boiling point of a ...

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

A

`-373.655K`

B

386.73K

C

392.86K

D

379.93K

Text Solution

Verified by Experts

The correct Answer is:
A
No of moles of NaCl `=("mass")/("molecular mass") = (3.5)/(58.5)`
So, actual number of moles of particle (ions) furnished by sodium chloride `=2 xx (3.5)/(58.5)`
Similarly, actual number of moles of particles (ions) furnished by magnesium chloride `=3 xx (0.13)/(95)`
Total number of moles of particles `= (2xx (3.5)/(58.5) + 3 xx (0.13)/(95))= 0.1238`
Mass of water `=(100-3.5-0.13)= 96.37g = (96.37)/(1000) kg`
Molality `=(0.1238)/(96.37)xx 1000= 1.2846`
`Delta T_(b)`= Molality `xx K_(b)`
`=1.2846 xx 0.51= 0.655K`
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