Question

MOLARITY REVIEW

1.) How many nitrate ions are present in a 25.00mL sample of a 2.50M calcium nitrate solution?

2.) How many mL of a 0.300M lead(II) nitrate solution are needed to react with 50.00mL of a 0.500M potassium chromate solution? (solution stoichiometry problem)

3.) How many mL of a 0.700M Ca(NO3)2 solution are necessary to react with 25.00mL of a 1.000M H2SO4 solution?

4.) 25.66mL of a solution hydroxide solution reacts with 22.55mL of a 0.888M phosphoric acid solution, determine the molarity (M) of the NaOH solution.

5.) A student added 25.00 grams of sodium nitrate to a 2.00L volumetric flask. What is the molarity of the resulting solution?

6.) A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00L volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3^-1?

7.) How many grams of phosphoric acid are required to react with 50.00mL of a 0.800M calcium nitrate solution?

Answer 1

Given data Volume of a sample of calcium nitrate solution, Moles of the solution. M1 2.50 M = 25.0 mL Write a chemical equation for the dissociation of the calcium nitrate Ca(NO32 (aqgCa(aa)+2NO; (aq) In this equation one mole of calcium nitrate produces two mole of nitrate ions Calculate the mole of nitrate ions in the sample =volume of sample (in liter) (V), × moles (14) of Ca (NO3)2 Substitute the values 0.025x 2.5 0.0625mol Calculate the number of nitrate ions mol of NO Number of NO3-Moles of Ca (NO3), X × Avogadro number mol of Ca(NO2 Substitute the values = 0.0625x-x 6.023×1023 -7.528x10" ions Hence, the number of nitrate ions present in sample is 7.528x1022

Given data: Volume of the potassium chromate solution, 50.0mL Molarity of the potassium chromate solution, M, 0.5M Molarity of the lead(II nitrate solution, M20.30M Let, the volume of the lead(II) nitrate be Write an expression to calculate volume of the lead(II) nitrate. Substitute the values 50.0x0.5 V,x0.3 83.33 mL Hence, the volume of the lead (II) nitrate required is 83.33 mL

Given data; Volume of the Phosphoric acid solution, 22.55mL Moles of the phosphoric acid solution, M 0.888M The volume of the NaOH solution. Ta: 25.66 mL Let, molarity of the NaOH solution be M2 Write a reaction of NaOH with phosphoric acid H3PO 3 NaOH + H,PO,-Na,PO, +3 H,O Write an expression to calculate moles of NaOH Substitute the values. 22.55x0.888 25.66x M2 M2 0.78M Three moles of NaOH are used to complete the reaction Calculate of molarity of NaOH = 3x 0.78 -2.34M Hence, the molarity of NaOH solution is 12.34 MI