.250 L of a .1 M solution contains .025 moles of solute
mass needed is ... .025 * 204.23 g
calculate the mass of potassium hydrogen phthalate needed to prepare a 0.1M solution in a 250 ml volumetric flask given that the molar mass of potassium hydrogen phthalate is 204,23g/mol
3 answers
204.23*250/1000*0.1=5.1058g
please asap...Potassium hydroxide reacts completely with sulfuric acid according to the equation:
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)
0.95 g of potassium hydroxide were dissolved in water and made up to 250 cm3 of solution. 25 cm3 samples of this solution were titrated against sulfuric acid (H2SO4) solution. The mean volume (titre) of sulfuric acid required for complete reaction was 15.00 cm3.
Find the molar concentration of the prepared solution of potassium hydroxide. Complete the calculation given below:
Relative Molecular Mass (RMM) of KOH
=
No. of moles of KOH in 250 cm3 =
=
molar concentration of KOH =
(Include units in your answer) =
No. of moles of KOH in 25.00 cm3 =
= ………………………… (1 mark)
no. of moles H2SO4 in 15.00 cm3 =
= ………………………… (1 mark)
molar concentration of H2SO4 =
(Include units in your answer) = ………………………… (2 marks)
(Total 10 marks)
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)
0.95 g of potassium hydroxide were dissolved in water and made up to 250 cm3 of solution. 25 cm3 samples of this solution were titrated against sulfuric acid (H2SO4) solution. The mean volume (titre) of sulfuric acid required for complete reaction was 15.00 cm3.
Find the molar concentration of the prepared solution of potassium hydroxide. Complete the calculation given below:
Relative Molecular Mass (RMM) of KOH
=
No. of moles of KOH in 250 cm3 =
=
molar concentration of KOH =
(Include units in your answer) =
No. of moles of KOH in 25.00 cm3 =
= ………………………… (1 mark)
no. of moles H2SO4 in 15.00 cm3 =
= ………………………… (1 mark)
molar concentration of H2SO4 =
(Include units in your answer) = ………………………… (2 marks)
(Total 10 marks)