34.4 g of aluminum nitrate was dissolved in 431.9 g of water. calculate the concentration in MOLALITY if the density of wter is 0.97 g/mL a. 0.374 m b. 0.363 m c. 0.852 m d. 0.534 m
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Given: Mass of aluminum nitrate (Al(NO3)3) = 34.4 g Molar mass of aluminum nitrate (Al(NO3)3) = 212.996 g/mol Number of moles of aluminum nitrate = Mass / Molar mass Number of moles of aluminum nitrate = 34.4 g / 212.996 g/mol Number of moles of aluminum nitrate Show more…
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Sima S.
Complete the following table for aqueous solutions of aluminum nitrate. $$ \begin{array}{cccc} \hline & \text { Mass of Solute } & \text { Volume of Solution } & \text { Molarity } \\ \hline \text { (a) } & 1.672 \mathrm{~g} & 145.0 \mathrm{~mL} & \\ \text { (b) } & 2.544 \mathrm{~g} & - & 1.688 \mathrm{M} \\ \text { (c) } & & 894 \mathrm{~mL} & 0.729 \mathrm{M} \end{array} $$
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