Question

The molal freezing point constant of water is $1.86C/M$. Therefore the freezing point of $0.1MNaCl$ solution in water is expected to be?

• A. $-{1.86}^{o}C$
• B. $-{0.186}^{o}C$
• C. $-{0.372}^{o}C$
• D. $+{0.372}^{o}C$

Answer 1

The correct option is C $-{0.372}^{o}C$

$△T_f=K_f\times m\times i$ $i=2$ (for $NaCl\to N{a}^{+}+C{l}^{-}$)

$M=0.1molal$
$K_f={1.86}^{o}C/molal$

$△T_f=I\times 1.86\times 0.1=0.186\times 2={0.372}^{o}C$

Therefore, now freezing point = $0-{0.372}^{o}C=-{0.372}^{o}C$
where i is vant hoff constant/factor.