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Q.
$15 \,g$ urea and $20 \,g \,NaOH$ dissolved in water. Total mass of solution is $250\, g$. Mole fraction of $NaOH$ in the mixture
Solutions
Solution:
mole fraction of $NaOH \Rightarrow \frac{\text { moles of } NaOH }{\text { moles of } NaOH +\text { moles of urea }+\text { moles of } H _{2} O }$
moles of $NaOH =\frac{ w }{ M _{ w }}=\frac{20}{40} \Rightarrow \frac{1}{2} $
$[ NaOH =23+16+1 \Rightarrow 40]$
moles of urea $\Rightarrow \frac{w}{M_{w}} \Rightarrow \frac{15}{60}$
moles of $H _{2} O \Rightarrow \frac{ w }{ M _{ w }} \Rightarrow \frac{215}{8}$
Mole fraction of $NaOH \Rightarrow \frac{\frac{1}{2}}{\frac{1}{2}+\frac{15}{60}+\frac{215}{18}} \Rightarrow 0.03$
