Ans.: (1000mL/1L)(1.18 g/mL)(0.37)(1mol HCl/36.5 g HCl) = 11.96 M
We want to make 100 mL of 6.0 M: 600 mmol HCl required
V x M = V x M 100 mL x 6.0 mmol/mL = V x 11.96 M
V = 50.17 mL Therefore, take 50 mL of the conc. HCl and dilute to exactly 100.0 mL with water.
Describe the preparation of 100 mL of 6.0 M HCL from a concentrated solution that has a specific gravity of 1.18 and is 37% (w/w) HCL
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tanx
πΆπ»πΆπ = 1.18 Γ103 π πππππππ‘
πΏ πππππππ‘ Γ 37 π π»πΆπ
100 π πππππππ‘ Γ 1 πππ π»πΆπ
36.5 π π»πΆπ = 12 π
The number of moles HCl required is given by
ππ. πππ π»πΆπ = 100 ππ Γ 1 πΏ
1000 ππΏ Γ 6 πππ π»πΆπ
πΏ = 0.600 πππ π»πΆπ
Finally, to obtain the volume of concentrated reagent we write
π½πππππ Γ πͺπππππ = π½π ππ Γ πͺπ ππ
ππππππ Γ 12 πππ
πΏ = 0.600 πππ π»πΆπ
ππππππ πππππππ‘ = 0.600 πππ π»πΆπ Γ 1 πΏππππππ π‘
12 π ππ π»πΆπ = 0.0500πΏ ππ 50.0 ππΏ
Therefore, dilute 50 mL of the concentrated reagent and complete the solution with 100mL water
πΏ πππππππ‘ Γ 37 π π»πΆπ
100 π πππππππ‘ Γ 1 πππ π»πΆπ
36.5 π π»πΆπ = 12 π
The number of moles HCl required is given by
ππ. πππ π»πΆπ = 100 ππ Γ 1 πΏ
1000 ππΏ Γ 6 πππ π»πΆπ
πΏ = 0.600 πππ π»πΆπ
Finally, to obtain the volume of concentrated reagent we write
π½πππππ Γ πͺπππππ = π½π ππ Γ πͺπ ππ
ππππππ Γ 12 πππ
πΏ = 0.600 πππ π»πΆπ
ππππππ πππππππ‘ = 0.600 πππ π»πΆπ Γ 1 πΏππππππ π‘
12 π ππ π»πΆπ = 0.0500πΏ ππ 50.0 ππΏ
Therefore, dilute 50 mL of the concentrated reagent and complete the solution with 100mL water
1 answer