Question

If oc‎ean wat‎er is co‎nsist of 3.5‎% (m‎/m) sa‎lt, i.‎e. assu‎ming Na‎Cl (MM ‎= 58.44 g), w‎hat wo‎uld be the fre‎ezing poi‎nt of an o‎cean? (Kf ‎H2O = 1.86°‎C m-1)

A. 0.0‎00°C

B. -1.1‎1°C

C. -2.‎23°C

D. -0.06‎51°C

Answer 1

NaCl dissociate into $Na^+\>Cl^-$

$-\Delta\>T_f=K_f.m.i$

Where $\Delta\>T_f=$ freezing point depression

$m=$ Molarity of the solution

$i=$ No. Of particles formed when compound dissolves

$3.5\%NaCl=\frac{Mass\>of\>NaCl}{Mass\>of\>solution}×100\%$

$3.5g$ of NaCl are dissolved in $96.5g$ of water

Gram of NaCl in $1$kg of water

$=\frac{1000}{96.5}×3.5$

$=36.269g$

Concentration of NaCl $=\frac{36.269}{58.44}=0.62062M$

$\Delta\>T_f=-1.86×0.62062×2$

$=-2.3087°C$

All options given are wrong.

Unless it is assumed percentage of salt is $3.5\%(m/v)$ not $(m/m)$

In such a case;

Mass of NaCl in $1kg$ of water

$=\frac{1000}{100}×3.5=35g$

Concentration $=\frac{35}{58.44}=0.5989$

$\Delta\>T_f=1.86×0.5989×2$

$=-2.23°C$