# Solutions, Acids, Bases Molarity of a Solution

1. 0.54 g of ammononium chlordie is dissolc=ved in 250 mL of solution. Calculate the molarity.

2. What volumne of solution would be required to deliver 6.84 g og 0.200 M aluminum sulfate solution?

3. Calculate the molarity of a solution if 126 mL of 6.25 M NaOH was diluted to a total volume of 250mL.

4. Calculate the pH of a solution with [H+] = 1.35 x 10-2(-2 is superscript)M. State whether the solution is an acid or a base. Why?

5. Calculate the hydorgen ion concentration of a colution with a pH of 2.4.

6. If 25 mL of a 0.45 M AgNO3 solution is added to an excess of a 1.2 M NaCl solution, what mass of AgCl will be preciptitated?

AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)

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1. 0.54 g of ammonium chloride is dissolved in 250 mL of solution. Calculate the molarity.

Solution: Molecular weight of Ammonium Chloride = 14+4+35.5 = 53.5 grams/mole

Number of moles of NH4Cl = 0.54/53.4 = 0.01011

Therefore Molarity = 0.01011/0.25 = 0.04M

2. What volume of solution would be required to deliver 6.84 g of 0.200 M aluminum sulfate solution?

Molarity = (Mass of ...

#### Solution Summary

The solution examines acids and bases to determine the molarity of a solution.