Question

In: Chemistry

1.a) Suppose a solution of 14.8M NH3. How many milliliters of this solution are required to...

1.a) Suppose a solution of 14.8M NH₃. How many milliliters of this solution are required to prepare 100 mL

of 1.0M NH₃, when diluted?

b) A stock solution of Nitric acid is 15.8 M. How many milliliters of the stock solution is required to make

up to 1.00 L of 0.12`M Nitric acid?

Solutions

Expert Solution

1.a)

Use the formula C₁V₁ =C₂V₂

Where, C₁ = initial concentration = 14.8M

V₁ = initial volume = ?

C₂ = final concentration = 1.0M

V₂ = final volume = 100 ml

C₁V₁ =C₂V₂ Thus

V₁ = C₂V₂ /C₁

Substitute the value in above equation

V₁ = 1.0100/14.8 = 6.7568 ml

6.7568 milliliterl of 14.8 M NH₃ stock solution required

Use the formula C₁V₁ =C₂V₂

Where, C₁ = initial concentration = 15.8M

V₁ = initial volume = ?

C₂ = final concentration = 0.12M

V₂ = final volume = 1.00 liter = 1000 ml

C₁V₁ =C₂V₂ Thus

V₁ = C₂V₂ /C₁

Substitute the value in above equation

V₁ = 0.121000/15.8 = 63.29 ml

63.29 milliliterl of 15.8 M Nitric acid stock solution required

queen_honey_blossom answered 1 year ago

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