Question

5.0 g of solid Al(NO3)3 was added to enough water to make a 100.0 mL solution. Al(NO3)3 is very soluble in water.

1. Write out the dissolution reaction that occurs when the solid Al(NO3)3 was added to water.

2. What is the concentration, in Molarity, of Al(NO3)3 once it is dissolved in the water?

3. If 53 mL of water was added to the original Al(NO3)3 solution described above, what is the new concentration in Molarity?

4. What is the concentration, in % w/v, of the solution created when 12.0 g of solid Al(NO3)3 was added to enough water to make a 150.0 mL solution?

Answer 1

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Al (NOs), dissolution AICNOS)s Art3 Caa) + 3 No3 (a9) CS) motarily no.of meles of salute (mot Volume of Solution (L) mass(9) molar mass(gh ! no-op males Motonly mass af Solute (9) molar mass af Saucte (g/md) Válunue of Sslution (L) Value of colution gfven data s Mass of A1Cno Motar mas af At(No3)3 a12.9 y/met 5.09 -1o00mL Da1L 5.09 a12.99 g/ms M = O-235 mdl/L X 0.I L O.235 M M,ViMaVformula Initial Here use we final M, Mofanity of itial Sol" 0.2usM Mdlanl af final sdlution= ?! Veluue of 100ml t53mL - t0omL V, Value of I/ 53 mL O.235M)(00 M2MIV M O.153 M O.I53 M 5.09 २२.११J/Mm० X O.IS3L OR M - mass of Al(No3 (9) Votune of Solution CmL) a.og