Question

4. 200 mL of a 5.6M solution to BaCl_2 have 750 mL of water added to it. What is the new concentration?

Answer 1

Answer:

4. 1.18 mol·L⁻¹

14. See below.

Explanation:

4. Dilution calculation

V₁c₁ = V₂c₂

Data:

V₁ = 200 mL; c₁ = 5.6 mol·L⁻¹

V₂ = 950 mL; c₂ = ?

Calculation:

c₂ = c₁ × V₁/V₂

c₂ = 5.6 mol·L⁻¹ × (200/950) = 1.18 mol·L⁻¹

The new concentration is 1.18 mol·L⁻¹ .

14. Boyle's Law graphs

We can write Boyle's Law as

pV = k or p = k/V or V= k/p

p and V are inversely related.

(a) As pressure increases, volume decreases. Thus, a graph of V vs p is a hyperbola.

(b) p = k/V =k(1/V)

 1/V = (1/k)p

   y  =   m x  + 0

A graph of 1/V vs p is a straight line.