Question

8C How many liters of a 3.14 M K2SO4 solution are needed to provide 81.3  g of K2SO4 (molar mass 174.01 g/mol)? Recall that M is equivalent to mol/L.

Expresss your answer to three significant figures.

7B Gastric acid pH can range from 1 to 4, and most of the acid is

HCl.

For a sample of stomach acid that is 1.26×10⁻²M in HCl, how many moles of HCl are in 12.1 mL of the stomach acid?

Express the amount to three significant figures and include the appropriate units.

PLEASE, IF YOU CAN, ANSWER ALL OF THEM I WOULD GREATLY APPRECIATE IT!

Answer 1

Answer – 8C) We are given, [K₂SO₄] = 3.14 M, mass of K₂SO₄ = 81.3 g

molar mass K₂SO₄ = 174.01 g/mol

We need to calculate moles of K₂SO₄

We know,

Moles of K₂SO₄ = mass of K₂SO₄ / molar mass of K₂SO₄

                           = 81.3 g / 174.01 g.mol^-1

                           = 0.467 moles

We know, molarity = moles / L

So, volume (L) = moles / molarity

                         = 0.467 moles / 3.14 M

                         = 0.149 L

So, 0.149 liters of a 3.14 M K₂SO₄ solution are needed to provide 81.3  g of K₂SO₄.

7 B) We are given, [HCl] = 1.26*10^-2 M , volume = 12.1 mL

We know,

Molarity = moles / L

So, moles of HCl = molarity of HCl * volume (L)

We need to convert the volume mL to L

We know,

1 mL = 0.001 L

So, 12.1 mL = ?

= 0.0121 L

So, moles of HCl = 1.26*10^-2 M * 0.0121 L

                         = 1.52*10^-4 moles

1.52*10^-4 moles of HCl are in 12.1 mL of the stomach acid.