Calculate the molar mass of solute in a solut | vedclass

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(D) The osmotic pressure formula is given by $\pi V = nRT = \frac{W_2}{M_2} RT$. Rearranging for molar mass $M_2$: $M_2 = \frac{W_2 RT}{\pi V}$. Given

Vedclass · Accepted Answer

$410 \ g \ mol^{-1}$

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(D) The osmotic pressure formula is given by $\pi V = nRT = \frac{W_2}{M_2} RT$. Rearranging for molar mass $M_2$: $M_2 = \frac{W_2 RT}{\pi V}$. Given: $W_2 = 1 \ g$,$V = 0.3 \ dm^3$,$\pi = 0.2 \ atm$,$T = 300 \ K$,$R = 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$. Substituting the values: $M_2 = \frac{1 \ g 	imes 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1} 	imes 300 \ K}{0.2 \ atm 	imes 0.3 \ dm^3}$. $M_2 = \frac{24.6}{0.06} \ g \ mol^{-1} = 410 \ g \ mol^{-1}$.

Calculate the molar mass of solute in a solution prepared by dissolving $1 \ g$ in $0.3 \ dm^3$ solvent having osmotic pressure $0.2 \ atm$ at $300 \ K$. $\left[R=0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1}\right]$

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