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(A) The formula for osmotic pressure is $\pi V = nRT = \frac{W_B}{M_B} RT$.Rearranging for molecular weight $M_B = \frac{W_B 	imes R 	imes T}{\pi

Vedclass · Accepted Answer

$1198$

Vedclass · Answer

(A) The formula for osmotic pressure is $\pi V = nRT = \frac{W_B}{M_B} RT$.Rearranging for molecular weight $M_B = \frac{W_B 	imes R 	imes T}{\pi 	imes V}$.Given: $W_B = 20 \ g$,$V = 500 \ mL = 0.5 \ L$,$T = 15 + 273 = 288 \ K$,$\pi = \frac{600}{760} \ atm$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.Substituting the values: $M_B = \frac{20 	imes 0.0821 	imes 288}{(600/760) 	imes 0.5}$.$M_B = \frac{20 	imes 0.0821 	imes 288 	imes 760}{600 	imes 0.5} \approx 1198 \ g \ mol^{-1}$.

$A$ solution is made by dissolving $20 \ g$ of a substance in $500 \ mL$ of water. Its osmotic pressure was found to be $600 \ mm$ of $Hg$ at $15 \ ^oC$. The molecular weight of the substance is

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