Asked by Density chemistry

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
Answered by john
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.
Answered by navoda
Good
Answered by sidh
tanx
Answered by Bassam
𝐢𝐻𝐢𝑙 = 1.18 Γ—103 𝑔 π‘Ÿπ‘’π‘Žπ‘”π‘’π‘›π‘‘
𝐿 π‘Ÿπ‘’π‘Žπ‘”π‘’π‘›π‘‘ Γ— 37 𝑔 𝐻𝐢𝑙
100 𝑔 π‘Ÿπ‘’π‘Žπ‘”π‘’π‘›π‘‘ Γ— 1 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
36.5 𝑔 𝐻𝐢𝑙 = 12 𝑀
The number of moles HCl required is given by
π’π‘œ. π‘šπ‘œπ‘™ 𝐻𝐢𝑙 = 100 π‘šπ‘™ Γ— 1 𝐿
1000 π‘šπΏ Γ— 6 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
𝐿 = 0.600 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
Finally, to obtain the volume of concentrated reagent we write
𝑽𝒄𝒐𝒏𝒄𝒅 Γ— π‘ͺ𝒄𝒐𝒏𝒄𝒅 = π‘½π’…π’Šπ’ Γ— π‘ͺπ’…π’Šπ’
π‘‰π‘π‘œπ‘›π‘π‘‘ Γ— 12 π‘šπ‘œπ‘™
𝐿 = 0.600 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
π‘‰π‘π‘œπ‘›π‘π‘‘ π‘Ÿπ‘’π‘Žπ‘”π‘’π‘›π‘‘ = 0.600 π‘šπ‘œπ‘™ 𝐻𝐢𝑙 Γ— 1 πΏπ‘Ÿπ‘’π‘Žπ‘”π‘’π‘› 𝑑
12 π‘š π‘œπ‘™ 𝐻𝐢𝑙 = 0.0500𝐿 π‘œπ‘Ÿ 50.0 π‘šπΏ
Therefore, dilute 50 mL of the concentrated reagent and complete the solution with 100mL water
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