Given:
Mixture of vitamin water = 75% pure water & 25% concentrated vitamin drink.
To find:
Quantity of water to be added to 16 gallons of concentrated vitamin drink to prepare a tank of vitamin water.
Solution:
Let total amount of mixture be x.
25% of x = 16 gallons (from given information)
When 25% of x is 16 gallons, the remaining 75% of x will be calculated as below:
(16/25)*75 = 48
Answer: 75% of x = 48 gallons. This means 48 gallons pure water is required to be added to mixture to prepare a tank of vitamin water.
Molar mass of vitamin B1, C12H17N4OS = 265.34 g/ mol
Molar mass of vitamin B2, C17H20N4O6 = 376.37 g/ mol
Molar mass of vitamin B5, C9H17NO5 = 219.24 g/ mol
Molar mass of vitamin B6, C8H11NO3 = 169.18 g/ mol
Molar mass of vitamin B7, C10H16N2O3S = 244.31 g/ mol
Now,
Order of increasing molar mass = B6 < B5 < B7 < B1 < B2
According to the reaction equation:
CH3COO- + H+ → CH3COOH
initial 0.25 0.15
change - 0.025 + 0.025
Equ (0.25-0.025) (0.15 + 0.025)
first, we have to get moles acetate and moles acetic acid:
moles of acetate = 0.25 - 0.025 = 0.225 moles
∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M
moles of acetic acid = 0.15 + 0.025 = 0.175 moles
∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M
Pka = -㏒ Ka
= -㏒ 1.8 x 10^-5
= 4.74
from H-H equation we can get the PH value:
PH = Pka + ㏒ [acetate / acetic acid]
PH = 4.74 + ㏒[0.225/0.175]
∴ PH = 4.8
It is caco3 ok thats the answer
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