If 20 g of a solute was dissolved in 500 mL | vedclass

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Question

(B) The formula for osmotic pressure is $\pi = \frac{n}{V}RT = \frac{m 	imes R 	imes T}{M 	imes V}$.Given:$m = 20 \ g$$V = 500 \ mL = 0.5 \ L$$\pi

Vedclass · Accepted Answer

$1200$

Vedclass · Answer

(B) The formula for osmotic pressure is $\pi = \frac{n}{V}RT = \frac{m 	imes R 	imes T}{M 	imes V}$.Given:$m = 20 \ g$$V = 500 \ mL = 0.5 \ L$$\pi = \frac{600}{760} \ atm$$T = 15 + 273 = 288 \ K$$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$Substituting the values:$\frac{600}{760} = \frac{20 	imes 0.0821 	imes 288}{M 	imes 0.5}$$M = \frac{20 	imes 0.0821 	imes 288 	imes 760}{600 	imes 0.5}$$M \approx 1200 \ g \ mol^{-1}$.

If $20 \ g$ of a solute was dissolved in $500 \ mL$ of water and the osmotic pressure of the solution was found to be $600 \ mm$ of $Hg$ at $15 \ ^oC$,then the molecular weight of the solute is:

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