Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol
3CaCl2+Al2O3 -------->3CaO +2AlCl3
mole from reaction 3 mol 2 mol
mass from reaction 3mol* 111.1g/mol 2 mol*133.5g/mol
333.3 g 267.0 g
mass from problem 45.7 g x g
Proportion:
333.3 g CaCl2 ------- 267.0 g AlCl3
45.7 g CaCl2 -------- x g AlCl3
x=45.7*267.0/333.3= 36.6 g AlCl3
Answer:
V2 = 35.967cm^3
Explanation:
Given data:
P1 = 0.2atm
P2 = 1.4atm
V1 = 250cm^3
V2 = ?
T1 = 10°C + 273 = 283K
T2 = 12°C + 273 = 285K
Apply combined law:
P1xV1/T1 = P2xV2/T2 ...eq1
Substituting values:
0.2 x 250/283 = 1.4 x V2/285
Solve for V2:
V2 = 14250/396.2
V2 = 35.967cm^3
