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Question Number 178165 by Spillover last updated on 13/Oct/22

$$2.86\mathrm{g}\mathrm{of}\mathrm{hydrated}\mathrm{Sodium}\mathrm{carbonate}$$$${\mathbf{Na}}_2{\mathbf{CO}}_3.{\mathbf{nH}}_2\mathbf{O}\mathrm{was}\mathrm{dissolved}$$$$\mathrm{into}\mathrm{water}\mathrm{to}\mathrm{make}{250\mathrm{cm}}^3\mathrm{solution}$$$$\mathrm{of}0.04\mathrm{M}\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{n}$$

Answered by a.lgnaoui last updated on 14/Oct/22

$$$$$${\mathrm{Na}}_2{\mathrm{CO}}_3+3\mathrm{H}2\mathrm{O}\to {2\mathrm{N}}_{\mathrm{a}}\mathrm{OH}+{\mathrm{CO}}_2+{2\mathrm{H}}_2\mathrm{O}$$$$1\mathrm{mole}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)\to 106g\to 3moles\left({\mathrm{H}}_2\mathrm{O}\right)$$$$2,86g\to N=\frac{m}{MV}=\frac{2,86}{106\times 250\times {10}^{-3}}=0,108moles\to 3\times 0,027=0,124moles\left({\mathrm{H}}_2\mathrm{O}\right)$$$$0,04moles\left({Na}_2{CO}_3\right)\to 0,108\times {10}^{-4}\times 3=0,344moles\left({\mathrm{H}}_2\mathrm{O}\right)=0,49\equiv \frac{1}{2}mole\left(H_{2}O\right)$$$$nousavons$$$$n=8({\mathrm{Na}}_2{\mathrm{CO}}_3,{8\mathrm{H}}_2\mathrm{O})?$$$$$$

Commented by Spillover last updated on 14/Oct/22

$$\mathrm{n}=10$$

Answered by Spillover last updated on 14/Oct/22

$$\mathrm{weight}\mathrm{of}\mathrm{hydrated}\left({\mathrm{w}}_1\right)=8.5\mathrm{g}$$$$\mathrm{weight}\mathrm{of}\mathrm{anhydrous}\left({\mathrm{w}}_2\right)=4.0\mathrm{g}$$$$\mathrm{molar}{\mathrm{mass}}_{{\mathbf{nH}}_2\mathbf{O}}=18\mathrm{n}$$$$\mathrm{molar}{\mathrm{mass}}_{{\mathbf{Na}}_2{\mathbf{CO}}_3}=160$$$$\frac{{\mathrm{w}}_1-{\mathrm{w}}_2}{{\mathrm{w}}_2}=\frac{18\mathrm{n}}{160}$$$$\frac{8.5-4.0}{4.0}=\frac{18\mathrm{n}}{160}$$$$\mathrm{n}=\frac{4.5\times 160}{4.0\times 18}$$$$\mathrm{n}=10$$

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