वाष्प दाब विधि में \(x_2=0.02\) मिला। यदि (0.18,g) विलेय (9,g) जल में घोला गया है, तो विलेय का मोलर द्रव्यमान लगभग क्या होगा?
In vapour pressure method, \(x_2=0.02\) is obtained. If (0.18,g) solute is dissolved in (9,g) water, what is the approximate molar mass of solute?
Explanation opens after your attempt
A. \(18,g,mol^{-1}\)
Concept
Moles of water \(=\frac{9}{18}=0.5\).
Why this answer is correct
\(x_2=\frac{n_2}{n_1+n_2}=0.02\), so \(n_2\approx\frac{0.02\times0.5}{0.98}\approx0.0102\) mol.
Exam Tip
Molar mass \(\frac{0.18}{0.0102}\approx18,g,mol^{-1}\). चरण 1: जल के मोल \(\frac{9}{18}=0.5\) हैं। चरण 2: \(x_2=\frac{n_2}{n_1+n_2}=0.02\), अतः \(n_2\approx\frac{0.02\times0.5}{0.98}\approx0.0102\) मोल। चरण 3: मोलर द्रव्यमान \(\frac{0.18}{0.0102}\approx18,g,mol^{-1}\) है।
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