(0.02,M) \(BaCl_2\) विलयन का (300,K) पर परासरण दाब कितना होगा, यदि वियोजन (75%) है और \(R=0.082,L,atm,mol^{-1},K^{-1}\)?
What will be the osmotic pressure of a (0.02,M) \(BaCl_2\) solution at (300,K), if dissociation is (75%) and \(R=0.082,L,atm,mol^{-1},K^{-1}\)?
Explanation opens after your attempt
C. (1.230,atm)
Concept
\(BaCl_2\) के लिए (i=1+\alpha(3-1)=1+0.75\times2=2.5)। / For \(BaCl_2\), (i=1+\alpha(3-1)=1+0.75\times2=2.5).
Why this answer is correct
\(\pi=iCRT=2.5\times0.02\times0.082\times300=1.230,atm\)। / \(\pi=iCRT=2.5\times0.02\times0.082\times300=1.230,atm\).
Exam Tip
परासरण दाब में आयनिक विलेय के लिए (i) अवश्य लगाएं। / Always include (i) for ionic solutes in osmotic pressure.
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