The solution containing 4.0 g of a polyvinyl | vedclass

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Question

(B) The formula for osmotic pressure is $\pi V = nRT = \frac{w}{M} RT$. Given: $\pi = 6.0 	imes 10^{-4} \ atm$,$V = 1 \ L$,$w = 4.0 \ g$,$T = 300 \ K

Vedclass · Accepted Answer

$1.6 	imes 10^{5}$

Vedclass · Answer

(B) The formula for osmotic pressure is $\pi V = nRT = \frac{w}{M} RT$. Given: $\pi = 6.0 	imes 10^{-4} \ atm$,$V = 1 \ L$,$w = 4.0 \ g$,$T = 300 \ K$,and $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$. Substituting the values: $6.0 	imes 10^{-4} 	imes 1 = \frac{4.0}{M} 	imes 0.082 	imes 300$. $M = \frac{4.0 	imes 0.082 	imes 300}{6.0 	imes 10^{-4}}$. $M = \frac{98.4}{6.0 	imes 10^{-4}} = 16.4 	imes 10^{4} = 1.64 	imes 10^{5} \ g \ mol^{-1}$. Thus,the molecular mass is approximately $1.6 	imes 10^{5} \ g \ mol^{-1}$.

The solution containing $4.0 \ g$ of a polyvinyl chloride polymer in $1 \ L$ of dioxane was found to have an osmotic pressure of $6.0 \times 10^{-4} \ atm$ at $300 \ K$. Given $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,the molecular mass of the polymer is:

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