The density of NaOH solution is 1.2 , g , cm^ | vedclass

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(B) Assume $1 \, L$ of the solution.Mass of solution $= 	ext{density} 	imes 	ext{volume} = 1.2 \, g \, cm^{-3} 	imes 1000 \, cm^3 = 1200 \, g$.Ass

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$5$

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(B) Assume $1 \, L$ of the solution.Mass of solution $= 	ext{density} 	imes 	ext{volume} = 1.2 \, g \, cm^{-3} 	imes 1000 \, cm^3 = 1200 \, g$.Assuming the molarity is $M$ (if not given,we typically assume a $1 \, M$ solution for such problems,or if the question implies $1 \, M$ based on the calculation steps provided in the original prompt).If we assume $1 \, M$ $NaOH$ solution:Moles of $NaOH = 1 \, mol$.Mass of $NaOH = 1 \, mol 	imes 40 \, g \, mol^{-1} = 40 \, g$.Mass of solvent (water) $= 1200 \, g - 40 \, g = 1160 \, g = 1.16 \, kg$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{1 \, mol}{1.16 \, kg} \approx 0.86 \, m$.However,based on the provided solution steps in the prompt which calculate $5 \, mol$ of $NaOH$ in $1 \, kg$ of water,the question implies a $5 \, M$ solution.For a $5 \, M$ solution:Moles of $NaOH = 5 \, mol$.Mass of $NaOH = 5 \, mol 	imes 40 \, g \, mol^{-1} = 200 \, g$.Mass of solution $= 1200 \, g$.Mass of solvent $= 1200 \, g - 200 \, g = 1000 \, g = 1 \, kg$.Molality $(m) = \frac{5 \, mol}{1 \, kg} = 5 \, m$.

The density of $NaOH$ solution is $1.2 \, g \, cm^{-3}$. The molality of this solution is $.... \, m$ (Round off to the nearest integer). [Use: Atomic masses: $Na: 23.0 \, u, O: 16.0 \, u, H: 1.0 \, u$]

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