A solution containing 4 g of a non-volatile | vedclass

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(B) The osmotic pressure formula is $\pi V = nRT = \frac{W_2}{Mw_2} RT$. Given: $\pi = \frac{500}{76} 	ext{ atm}$,$V = 100 \ mL = 0.1 \ L$,$W_2 = 4 \

Vedclass · Accepted Answer

$149.7$

Vedclass · Answer

(B) The osmotic pressure formula is $\pi V = nRT = \frac{W_2}{Mw_2} RT$. Given: $\pi = \frac{500}{76} 	ext{ atm}$,$V = 100 \ mL = 0.1 \ L$,$W_2 = 4 \ g$,$R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,$T = 27 + 273 = 300 \ K$. Substituting the values: $\frac{500}{76} 	imes 0.1 = \frac{4}{Mw_2} 	imes 0.0821 	imes 300$. $Mw_2 = \frac{4 	imes 0.0821 	imes 300 	imes 76}{500 	imes 0.1}$. $Mw_2 = \frac{7475.52}{50} = 149.51 \approx 149.7$.

$A$ solution containing $4 \ g$ of a non-volatile organic solute per $100 \ mL$ has an osmotic pressure of $500 \ cm$ of mercury at $27^\circ C$. What is the molar mass of the solute?

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