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The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
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Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.
Explanation :
Let the volume of sodium benzoate (salt) be, x
So, the volume of benzoic acid (acid) will be, (100 - x)
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
x = 29.0
The volume of sodium benzoate = x = 29.0 mL
The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL
Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.
