Answer:
130ml of HCl(36%) in 4.90L solution => pH = 1.50
Explanation:
Need 4.90L of HCl(aq) solution with pH = 1.5.
Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺
[HCl(36%)] ≅ 12M in HCl
(M·V)concentrate = (M·V)diluted
12M·V(conc) = 0.032M·4.91L
=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.
Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.
Using ideal gas equation,
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=1 atm
T=25 C+273 K =298.15K
V=663 ml=0.663L
R=0.0821 atm L mol ⁻¹
Mass of gas given=1.25 g g
Molar mass of gas given=?
Putting all the values in the above equation,
Molar mass of the gas=46.15
Answer:
C = 0.08M
Explanation:
molar mass of AlCl3
Al =27
Cl = 35.5
27+3(35.5) =133.5g/mol
n= mass/Molar mass
n =CV
CV = mass/molar mass
C x 500 x 10^-³ = 5/133.5
C x 500 x 10^-³ = 0.04
C = 0.04/500 x 10^-³
C = 0.08M
