At a temperature of 27,^oC,0.6,g of a solute | vedclass

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(B) The formula for osmotic pressure is $\pi V = nRT$,where $n = \frac{w}{M}$.Rearranging for molar mass $M$,we get $M = \frac{wRT}{\pi V}$.Given: $w

Vedclass · Accepted Answer

$120$

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(B) The formula for osmotic pressure is $\pi V = nRT$,where $n = \frac{w}{M}$.Rearranging for molar mass $M$,we get $M = \frac{wRT}{\pi V}$.Given: $w = 0.6\,g$,$T = 27 + 273 = 300\,K$,$R = 0.082\,L\,atm\,K^{-1}\,mol^{-1}$,$\pi = 1.23\,atm$,and $V = 0.1\,L$.Substituting the values: $M = \frac{0.6 	imes 0.082 	imes 300}{1.23 	imes 0.1}$.$M = \frac{14.76}{0.123} = 120\,g\,mol^{-1}$.

At a temperature of $27\,^oC$,$0.6\,g$ of a solute is dissolved in $0.1\,L$ of solvent,resulting in an osmotic pressure of $1.23\,atm$. The molar mass of the solute is .......... $g\,mol^{-1}$.

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