A 0.200 M NaOH solution was used to titrate a 18.25 mL HE solution. The endpoint was reached after 31.20 mL of titrant
were added. Find the molar concentration of the original HF
solution.

M₂ = 0.34 M

Explanation:

Given data:

Molarity of NaOH =M₁ =  0.200 M

Volume of HF =V₂=  18.25 mL

Volume of NaOH added = V₁ = 31.20 mL

Molarity of HF solution = M₂ = ?

Solution:

M₁V₁  =  M₂V₂

by putting values,

0.200 M × 31.20 mL = M₂ × 18.25 mL

M₂ = 0.200 M × 31.20 mL /  18.25 mL

M₂ = 6.24 M.mL / 18.25 mL

M₂ = 0.34 M

organisms are more likely to be more preserved in mud than sand because mud is moist and wet, while sand is very dry.