Molarity (M) of an aqueous solution containin | vedclass

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(B) Given: Molarity $(M) = 0.2 \ M$,Volume $(V) = 500 \ mL = 0.5 \ L$,Density $(d) = 1.25 \ g/mL$,Molar mass of $CuSO_4 = 159.5 \ g/mol$.Step $1$: Cal

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$164$

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(B) Given: Molarity $(M) = 0.2 \ M$,Volume $(V) = 500 \ mL = 0.5 \ L$,Density $(d) = 1.25 \ g/mL$,Molar mass of $CuSO_4 = 159.5 \ g/mol$.Step $1$: Calculate moles of solute $(n)$:$n = M 	imes V = 0.2 	imes 0.5 = 0.1 \ mol$.Step $2$: Calculate mass of solute $(x)$:$x = n 	imes 	ext{Molar mass} = 0.1 	imes 159.5 = 15.95 \ g$.Step $3$: Calculate mass of solution:$Mass_{soln} = V 	imes d = 500 	imes 1.25 = 625 \ g$.Step $4$: Calculate mass of solvent $(W_{solvent})$:$W_{solvent} = Mass_{soln} - x = 625 - 15.95 = 609.05 \ g$.Step $5$: Calculate molality $(m)$:$m = \frac{n 	imes 1000}{W_{solvent} (in \ g)} = \frac{0.1 	imes 1000}{609.05} \approx 0.16418 \ m$.$m \approx 164 	imes 10^{-3} \ m$.

Molarity $(M)$ of an aqueous solution containing $x \ g$ of anhydrous $CuSO_4$ in $500 \ mL$ solution at $32^{\circ} C$ is $2 \times 10^{-1} \ M$. Its molality will be. . . . . .$\times 10^{-3} \ m$ (nearest integer). [Given density of the solution $= 1.25 \ g/mL$.]

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