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Question

(B) Moles of solute '$A$' = $\frac{0.3 \ g}{60 \ g \ mol^{-1}} = 0.005 \ mol$.Moles of solute '$B$' = $\frac{0.9 \ g}{180 \ g \ mol^{-1}} = 0.005 \ mo

Vedclass · Accepted Answer

$2.46$

Vedclass · Answer

(B) Moles of solute '$A$' = $\frac{0.3 \ g}{60 \ g \ mol^{-1}} = 0.005 \ mol$.Moles of solute '$B$' = $\frac{0.9 \ g}{180 \ g \ mol^{-1}} = 0.005 \ mol$.Total moles of solute = $0.005 + 0.005 = 0.01 \ mol$.Volume of solution = $100 \ mL = 0.1 \ L$.Molarity $(C)$ = $\frac{	ext{Total moles}}{	ext{Volume in } L} = \frac{0.01 \ mol}{0.1 \ L} = 0.1 \ M$.Temperature $(T)$ = $27 + 273 = 300 \ K$.Using the formula for osmotic pressure: $\pi = CRT$.$\pi = 0.1 \ mol \ L^{-1} 	imes 0.082 \ L \ atm \ K^{-1} \ mol^{-1} 	imes 300 \ K$.$\pi = 2.46 \ atm$.

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