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ella [17]
3 years ago
10

You are starting with 2 pounds of 30% acetic acid solution. You want to dilute

Chemistry
1 answer:
Tpy6a [65]3 years ago
8 0

Answer:

The mass of water to be added is 2 pounds

Explanation:

The given parameters are;

The mass of the given solution = 2 pounds

The concentration of the given solution = 30%

The desired concentration of the solution = 15%

The mass, m of the acetic acid in the given solution = 30% × 2 pounds

m = 30/100 × 2 pounds = 0.6 pounds

To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;

15% of X = 0.6 pounds

15/100 × X = 3/20 × 0.6 pounds

∴ The mass of the solution required  X = 0.6 × 20/3 = 4 pounds

The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds

Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.

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2 years ago
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P_1=0.85atm\\V_1=?\\T_1=66^oC=[66+273]K=339K\\P_2=0.60atm\\V_2=94mL\\T_2=43^oC=[43+273]K=316K

Now put all the given values in above equation, we get:

\frac{0.85atm\times V_1}{339K}=\frac{0.60atm\times 94mL}{316K}

V_1=71.2mL

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