Answer:
A. 0.2395 w/w %
B. 2394ppm
Explanation:
A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:
Mass glycerol / Total mass * 100
<em>Mass glycerol:</em>
The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):
2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol
<em>Mass of water:</em>
998.9mL and density = 0.9982g/mL:
998.9mL * (0.9982g/mL) = 997.1g of water.
That means percent by mass is:
% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %
B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:
2.394g * (1000mg / 1g) = 2394mg:
Parts per million: 2394mg / L = 2394ppm
0,15 moles of NaOH-------in------------1000ml
x moles of NaOH------------in--------100ml
x = 0,015 moles of NaOH
final volume = 150ml
0,015 moles of NaOH---in-------150ml
x moles of NaOH--------------in-----1000ml
x = 0,1 moles of NaOH
answer: 0,1mol/dm³ (molarity)
Answer: 0.024 M
Explanation:
According to the neutralization law,
where,
= molarity of stock solution = 0.12 M
= volume of stock solution = 20.0 ml
= molarity of diluted solution = ? M
= volume of diluted solution = 100.0 ml
Putting in the values we get:
Therefore, the concentration of the resulting solution is 0.024 M.
