A centimolar solution of acetic acid is 50 % | vedclass

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

Question

(A) For a centimolar solution,the concentration $C = 0.01 \ M = 10^{-2} \ mol \ L^{-1}$.Acetic acid $(CH_3COOH)$ dissociates as: $CH_3COOH \rightlefth

Vedclass · Accepted Answer

$0.37$

Vedclass · Answer

(A) For a centimolar solution,the concentration $C = 0.01 \ M = 10^{-2} \ mol \ L^{-1}$.Acetic acid $(CH_3COOH)$ dissociates as: $CH_3COOH ightleftharpoons CH_3COO^- + H^+$.The degree of dissociation $\alpha = 0.5$.The van't Hoff factor $i = 1 + \alpha(n-1)$,where $n=2$.$i = 1 + 0.5(2-1) = 1.5$.The osmotic pressure $\pi = iCRT$.Given $T = 27 + 273 = 300 \ K$ and $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$.$\pi = 1.5 	imes 0.01 	imes 0.083 	imes 300$.$\pi = 1.5 	imes 0.01 	imes 24.9 = 1.5 	imes 0.249 = 0.3735 \ atm$.Rounding to the nearest option,the value is $0.37 \ atm$.

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

Similar Questions

Two solutions of $KNO_3$ and $CH_3COOH$ are prepared separately. Molarity of both is $0.1 \ M$ and osmotic pressures are $P_1$ and $P_2$ respectively. The correct relationship between the osmotic pressures is :-

Elevation in boiling point of the aqueous solution of $0.01 \ M \ BaCl_2$ compared to $0.01 \ M$ urea is . . . . . . .

In equimolar solutions of glucose,$NaCl$,and $BaCl_2$,the order of osmotic pressure is as follows:

Van't Hoff factor of $Hg_{2}Cl_{2}$ in its aqueous solution will be : ($Hg_{2}Cl_{2}$ is $80\%$ ionized in the solution) :- (in $.6$)

When mercuric iodide $(HgI_2)$ is added to an aqueous solution of potassium iodide $(KI)$,the

Vedclass Test Series

Exam Paper Generator

Online Exam Module