8C How many liters of a 3.14 M K2SO4 solution are needed to provide 81.3 g of K2

ID: 973822 • Letter: 8

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 1.26×102M 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 – 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.

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