The osmotic pressure of a solution of NaCl is | vedclass

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(B) Osmotic pressure $\pi = i 	imes C 	imes RT$ For $NaCl$,$i = 2$. Given $\pi_{NaCl} = 0.10 \ atm = 2 	imes C_{NaCl} 	imes RT$. Thus,$C_{NaCl}

Vedclass · Accepted Answer

$167$

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(B) Osmotic pressure $\pi = i 	imes C 	imes RT$ For $NaCl$,$i = 2$. Given $\pi_{NaCl} = 0.10 \ atm = 2 	imes C_{NaCl} 	imes RT$. Thus,$C_{NaCl} 	imes RT = 0.05 \ atm$. For glucose,$i = 1$. Given $\pi_{glucose} = 0.20 \ atm = 1 	imes C_{glucose} 	imes RT$. Thus,$C_{glucose} 	imes RT = 0.20 \ atm$. Number of moles of $NaCl$ in $1 \ L$ is $n_{NaCl} = C_{NaCl} 	imes 1 = \frac{0.05}{RT}$. Number of moles of glucose in $2 \ L$ is $n_{glucose} = C_{glucose} 	imes 2 = \frac{0.20 	imes 2}{RT} = \frac{0.40}{RT}$. Total volume $V_{total} = 1 \ L + 2 \ L = 3 \ L$. Total osmotic pressure $\pi_{total} = \frac{(n_{NaCl} 	imes i_{NaCl} + n_{glucose} 	imes i_{glucose}) 	imes RT}{V_{total}}$. $\pi_{total} = \frac{(\frac{0.05}{RT} 	imes 2 + \frac{0.40}{RT} 	imes 1) 	imes RT}{3} = \frac{0.10 + 0.40}{3} = \frac{0.50}{3} \ atm$. $\pi_{total} = 0.1666... \ atm = 166.6... 	imes 10^{-3} \ atm$. Rounding to the nearest integer,$x = 167$.

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