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A liquid mixture containing 40.0 wt% *n*-octane and the balance *n*-decane flows into a tank mounted on a balance. The mass in kg indicated by the scale is plotted against time. The data fall on a straight line that passes through the points (*t *= 3 min, *m *= 150 kg) and (*t*= 10 min, *m *= 250 kg).

(a) Estimate the volumetric flow rate of the liquid mixture.

(b) What does the empty tank weigh?

.

**(a)**

**Interpretation:**

Volumetric flow rate of the mixture is to be calculated.

**Concept introduction:**

Volumetric flow rate of a fluid is the volume of a fluid that passes per unit time. It is mathematically represented as follows:

Volumetric flow rate =

Here,

## Answer to Problem 5.1P

19.85 m^{3}/min

### Explanation of Solution

Given Information: Density of n-Octane = _{1} = 703 kg/m^{3} Density of n-Decane = _{2} = 730 kg/m^{3} Mole fraction of n-Octane = x_{1} = 0.4 Mole fraction of n-Decane = x_{2} = 0.6

Average density of the mixture = _{avg}

^{3} =719.2 kg/m^{3}

Given graph is a straight line with slope as follows:

=

Putting the values, =

=19.85 m^{3}/min

**(b)**

**Interpretation:**

Weight of empty tank is to be calculated.

**Concept introduction:**

A straight line equation is y = mx + c, where,

y = y-axis coordinate

x = x- axis coordinate

m = slope of line

c = intercept of straight line

Weight of a substance = Mass of substance

## Answer to Problem 5.1P

1050.16 N

### Explanation of Solution

The equation is a straight line with equation

m = at + c

where,

m = mass

a = slope of line

t = time

c = intercept

Mass of empty tank is mass at t = 0

i.e.

m = 0 + c

Mass of empty tank = intercept of the straight line

a =

a = 14.28 kg/min

At t = 3 min, m =150 kg

So,

150 = (14.28) (3) + c

c = 107.16

Therefore, mass of empty tank = 107.16 kg

Now, weight can be calculated using the following formula:

Here, g is acceleration due to gravity.

Putting the values,

W = 107.16 ^{2} = 1050.16 N

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Elementary Principles of Chemical Processes, Binder Ready Version

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