At 300 K, the osmotic pressure of a decinorm | vedclass

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

Question

(B) The osmotic pressure formula is given by $\pi = iCRT$. Given: $\pi = 4.82 \ atm$, $C = 0.1 \ M$ (for $NaCl$, $N = M$), $R = 0.082 \ L \ atm \ K^{-

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(B) The osmotic pressure formula is given by $\pi = iCRT$. Given: $\pi = 4.82 \ atm$, $C = 0.1 \ M$ (for $NaCl$, $N = M$), $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$, and $T = 300 \ K$. Calculating the van't Hoff factor $(i)$: $i = \frac{\pi}{CRT} = \frac{4.82}{0.1 	imes 0.082 	imes 300} = \frac{4.82}{2.46} \approx 1.959 \approx 1.96$. For $NaCl$, the dissociation is $NaCl ightarrow Na^+ + Cl^-$, so $n = 2$. The degree of dissociation $(\alpha)$ is related to $i$ by $i = 1 + \alpha(n - 1)$. $1.96 = 1 + \alpha(2 - 1) \Rightarrow \alpha = 0.96$. Since $\alpha = x 	imes 10^{-2}$, we have $0.96 = x 	imes 10^{-2}$, which gives $x = 96$.

At $300 \ K$, the osmotic pressure of a decinormal solution of sodium chloride is $4.82 \ atm$. The degree of dissociation of sodium chloride is $x \times 10^{-2}$. The value of $x$ is $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

Similar Questions

The van't Hoff factor $(i)$ is .....

If the degree of dissociation of an aqueous solution of a weak monobasic acid is determined to be $0.3$,then the observed freezing point depression will be $.....\%$ higher than the expected/theoretical freezing point depression. (Nearest integer)

What is association? What is dissociation?

$A$ solute $A$ undergoes association in a medium as $nA \rightleftharpoons A_n$. If the degree of association is $\alpha$,then the van't Hoff factor $i$ is given by:

The experimentally determined molar mass of a non-volatile solute, $BaCl_2$, in water by Cottrell's method, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module