1) The formula weight of the solid, monoprotic acid, \"KHP\" is 204.2 how many m

Question

1) The formula weight of the solid, monoprotic acid, "KHP" is 204.2 how many moles are present in 0.479g?

2) If 25.21 mL of NaOH solution is required to react with 0.555g of KHP, what is the molarity of the NaOH solution?

3) The titration of an impure sample of KHP found that 38.0mL of 0.100M NaOH was required to react completely with o.768g of sample. What is the percentage of KHP in this sample?

4) How many mL of 0.250M NaOH are required to neutralize 30.2mL of 0.149 M HCL?

Please explain, thank you

Explanation / Answer

1) moles of KHP = grams/molar mass = 0.479 g/204.2 g/mol = 0.0023 mol

2) moles of KHP = 0.555 g/204.2 g/mol = 0.00272 mol

Molarity of NaOH = moles/volume (L) = 0.00272 mol/0.02521 L = 0.108 M

3) moles of NaOH used = 0.1 M x 0.038 L = 0.0038 mol

actual mass of KHP in sample = 0.0038 mol x 204.2 g/mol = 0.776 g

wt% of KHP in sample = (0.776g/0.768g) x 100 = 101.036%

[pl. check your sample mass, I believe the number reported is incorrect]

4) moles of HCl = 0.149 M x 30.2 ml = 4.4998 mol

ml of NaOH reqired = mol/molarity = 4.4998 mol/0.250 M = 17.9992 ml or 18.0 ml

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