Question

Van't Hoff factor, (i), of a $0.5\%$ (w/W) aqueous solution of $KCl$ which freezes at $-{0.24}^o$C is:

($K_f$ of water $=1.86$ K kg $mo{l}^{-1}$, molecular weight of $KCl$$=74.5$)

• A. $1.52$
• B. $2.32$
• C. $1.92$
• D. $1.32$

Answer 1

Answer: (C)

$1.92$

We know that

$△T_b=i\times K_b\times molality$

$0-(-0.24)=i\times 1.86\times \frac{w}{MolecularWeight}\times 1000$

$0.24=i\times 1.86\times \frac{0.5}{74.5\times 100}\times 1000$

$i=1.92$