For 0.01 , M solutions of urea, common salt ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The depression of freezing point $(\Delta T_f)$ is given by the formula $\Delta T_f = i 	imes K_f 	imes m$, where $i$ is the van't Hoff factor.S

Vedclass · Accepted Answer

$1 : 2 : 3$

Vedclass · Answer

(C) The depression of freezing point $(\Delta T_f)$ is given by the formula $\Delta T_f = i 	imes K_f 	imes m$, where $i$ is the van't Hoff factor.Since the molality $(m)$ and the freezing point depression constant $(K_f)$ are the same for all solutions, the ratio of $\Delta T_f$ is equal to the ratio of their van't Hoff factors $(i)$.For urea (a non-electrolyte), $i = 1$.For common salt $(NaCl)$, it dissociates as $NaCl ightarrow Na^+ + Cl^-$, so $i = 2$.For $Na_2SO_4$, it dissociates as $Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$, so $i = 3$.Therefore, the ratio of depression of freezing point is $1 : 2 : 3$.

For $0.01 \, M$ solutions of urea, common salt $(NaCl)$, and $Na_2SO_4$, the ratio of depression of freezing point is:

Similar Questions

$A$ centimolar solution of acetic acid is $50 \%$ dissociated at $27^{\circ} C$. The osmotic pressure of the solution (in $atm$) is $\left(R=0.083 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

The freezing point of an equimolal aqueous solution will be highest for:

For a $0.1 \ M \ Ba(NO_3)_2$ solution, the van't Hoff factor is $2.74$. The degree of dissociation is .............

For a $0.25 \ m$ solution of $SrCl_2$,the van't Hoff factor is found to be $2.5$. What will be the effective molality of the solution in $m$?

What is the expected value of $\Delta T_f$ for $1.25 \ m$ $CaCl_2$ solution if $1.25 \ m$ sucrose solution has $\Delta T_f$ value $x \ K$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module