An inhibitor, which slows down the reaction enough to measure the release of gas.
Q1)
molarity is defined as the number of moles of solute in 1 L of solution.
the NaCl solution volume is 1.00 L
number of moles NaCl = NaCl mass present / molar mass of NaCl
NaCl moles = 112 g / 58.5 g/mol = 1.91 mol
the number of moles of NaCl in 1.00 L of solution is - 1.91 mol
therefore molarity of NaCl is 1.91 M
Q2)
molality is defined as the number of moles of solute in 1 kg of solvent.
density is mass per volume.
density of the solution is 1.08 g/mL.
therefore mass of the solution is = density x volume
mass = 1.08 g/mL x 1000 mL = 1080 g
since we have to find the moles in 1 kg of solvent
mass of solvent = 1080 g - 112 g = 968 g
number of moles of NaCl in 968 g of solvent - 1.91 mol
therefore number of NaCl moles in 1000 g - (1.91 mol / 968 g) x 1000 g/kg = 1.97 mol/kg
molality of NaCl solution is 1.97 mol/kg
Q3)
mass percentage is the percentage of mass of solute by total mass of the solution
mass percentage of solution = mass of solute / total mass of the solution
mass of solute = 112 g
total mass of solution = 1080 g
mass % of NaCl = 112 g / 1080 g x 100%
therefore mass % of NaCl = 10.4 %
answer is 10.4 %
The correct answer is A, since it is the only sensible answer.
The correct answer would b C
Answer:
Molarity of Sr(OH)₂ = 0.47 M
Explanation:
Given data:
Volume of Sr(OH)₂ = 15.0 mL
Volume of HCl = 38.5 mL (0.0385 L)
Molarity of HCl = 0.350 M
Concentration/Molarity of Sr(OH)₂ = ?
Solution:
Chemical equation:
Sr(OH)₂ + 2HCl → SrCl₂ +2H₂O
Number of moles of HCl:
Molarity = number of moles/ volume in L
0.350 M = number of moles/0.0385 L
Number of moles = 0.350 mol/L× 0.0385 L
Number of moles = 0.0135 mol
Now we will compare the moles of HCl with Sr(OH)₂.
HCl : Sr(OH)₂
2 : 1
0.0135 : 1/2×0.0135 = 0.007 mol
Molarity/concentration of Sr(OH)₂:
Molarity = number of moles / volume in L
Molarity = 0.007 mol /0.015 L
Molarity = 0.47 M
