What is the molar mass of a solute if a solut | vedclass

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(A) The formula for osmotic pressure is $\pi = CRT = \frac{n}{V} RT = \frac{w}{M_B 	imes V} RT$. Given: $\pi = 0.25 \ atm$,$w = 4.5 \ g$,$V = 3 \ dm^

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$148 \ g \ mol^{-1}$

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(A) The formula for osmotic pressure is $\pi = CRT = \frac{n}{V} RT = \frac{w}{M_B 	imes V} RT$. Given: $\pi = 0.25 \ atm$,$w = 4.5 \ g$,$V = 3 \ dm^3$,$T = 300 \ K$,and $R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$. Substituting the values: $0.25 = \frac{4.5}{M_B 	imes 3} 	imes 0.0821 	imes 300$. $0.25 = \frac{4.5 	imes 0.0821 	imes 100}{M_B}$. $M_B = \frac{4.5 	imes 8.21}{0.25} = 18 	imes 8.21 = 147.78 \ g \ mol^{-1}$. Rounding to the nearest integer,$M_B \approx 148 \ g \ mol^{-1}$.

What is the molar mass of a solute if a solution is prepared by dissolving $4.5 \ g$ of solute in $3 \ dm^3$ of water having an osmotic pressure of $0.25 \ atm$ at $300 \ K$? $(R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1})$

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