1) Molarity
M = n / V
n: number of moles of solute
V: volume of the solution in liters
n = mass / molar mass = 0.000333 g / 332.32 g / mol = 1*10 ^ - 6 moles
V = 225 ml * 1 liter / 1000 ml = 0.225 liter
M = 10^-6 mol / 0.225 liter = 0.00000444 M
2) ppm
ppm = parts per million
grams of solute: 0.000333 g
grams of solution = volume * density = 225 ml * 0.785 g / ml = 176.625 g
ppm = [0.00033 g / 176.625 g] * 1,000,000 = 1.868 ppm
5g NaOH x 1 mol NaOH/ 39.99g NaOH = 0.125 mol NaOH
Answer:
The answer to your question is 900 g of water vapor
Explanation:
Data
mass of H₂O = ?
mass of butane = 580 g
Balanced chemical reaction
2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O
Process
1.- Calculate the molar weight of butane and water
Butane (C₄H₁₀) = 2[(12 x 4) + (1 x 10)]
= 2[48 + 10]
= 2[58]
= 116 g
Water (H₂O) = 10[(1 x 2) + (1 x 16)]
= 10[2 + 16]
= 10[18]
= 180 g
2.- Use proportions and cross multiplication to find the mass of water vapor
116 g of butane ------------- 180 g of water
580 g of butane ---------- x
x = (580 x 180) / 116
x = 900 g of water vapor
