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Question #2177b

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Jul 18, 2017

$χ_{N a O H} ≅ 0.05$ $chi_(NaOH)~=0.05$

Explanation:

Now..... $Molality$ $"Molality"$ $=$ $=$ $\frac{Moles of solute}{Kilograms of solvent.}$ $"Moles of solute"/"Kilograms of solvent."$

And our $N a O H (a q)$ $NaOH(aq)$ solution is $3.0 \cdot molal$ $3.0*"molal"$ , i.e. $3.0 \cdot m o l \cdot k g^{- 1}$ $3.0*mol*kg^-1$

Now $χ_{N a O H} = \frac{Moles of NaOH}{Moles of NaOH + moles of water}$ $chi_(NaOH)="Moles of NaOH"/"Moles of NaOH + moles of water"$

$= \frac{3.0 \cdot m o l}{3.0 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o l^{- 1}}}$ $=(3.0*mol)/(3.0*mol+(1000*g)/(18.01*g*mol^-1))$

$= \frac{3.0 \cdot m o l}{3.0 \cdot m o l + 55.52 \cdot m o l} = 0.0513$ $=(3.0*mol)/(3.0*mol+55.52*mol)=0.0513$

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