Question

The freezing point of 1 m NaCl solution assuming NaCl to be 100% dissociated in water is ($K_f=1.86K.m^{-1}$)

• A. $-{1.86}^{o}C$
• B. $-{3.72}^{o}C$
• C. $+{1.86}^{o}C$
• D. $+{3.72}^{o}C$

Answer 1

Answer: (B)

$-{3.72}^{o}C$

$\Delta T_f=i\times K_f\times m=2\times 1.86\times 1=3.72K={3.72}^{o}C$;

$\therefore T_f=0-3.72=-{3.72}^{o}C$

NaCl dissociates to give experimental molality,
$i=1\times 2=2.$

Option B is correct.