Question

a.Gastric acid pH can range from 1 to 4, and most of the acid is HCl.

For a sample of stomach acid that is 2.50×10⁻²M in HCl, how many moles of HCl are in 11.9 mL of the stomach acid?

b.Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it. The median lethal dose of KCN for a person weighing 195 lb (88.5 kg ) is 7.47×10⁻³ mol .

What volume of a 0.0210 M KCN solution contains 7.47×10⁻³ mol of KCN?

c.. A 631 mL NaCl solution is diluted to a volume of 1.02 L and a concentration of 6.00 M . What was the initial concentration?

Answer 1

a)

0.0002975 moles

Explanation

Molarity is defined as the number of moles of solute per liter of solution

2.50×10^-2M HCl = 2.50 ×10^-2 moles of HCl per liter of solution

Therefore,

Number of moles of HCl in 11.9ml of stomach acid = (2.50×10^-2mol/1000ml) × 11.9ml = 0.0002975mol

b)

355.7ml

Explanation

0.0210M KCN = 0.0210 moles of KCN in 1000ml of the solution

Therefore

Volume of KCN = (1000ml / 0.0210 mol ) × 7.47×10^-3 mol = 355.7ml

c)

9.70M

Explanstion

dilution formula is as follows

C1× V1 = C2 × V2

C1 = Concentration of Initial solutiin

V1 = Volumw of Initial solution

C2 = Concentration of final solution

V2 = Volume of final solution

C1 = C2 × V2/V1

C1 = 6.00M × 1.02L / 0.631M = 9.70M