You mix 325 gallons of a mixture containing 4 wt% NaCI (sodium chloride) and 96

Question

You mix 325 gallons of a mixture containing 4 wt% NaCI (sodium chloride) and 96 wt% water (solution density = 1.0253 g/cm^3) with a unknown volume of solution that is 26 wt% NaCI and 74 wt% water (solution density = 1.1944 g/cm^3). You want to produce a mixture containing 15 wt% NaCI. Find the volume of the second feed solution by following the steps below. Label the flow chart of the mixing process Drag the labels below to fill in the spaces next to each quantity. If you are not given a value for a quantity, place the word 'unknown" in the space. DO NOT LEAVE ANY SPACES EMPTY. How many unknowns are in the problem statement Which independent equations can you write to solve for the volume of the second feed solution. Check all that apply. What is the volume of the second feed solution

Explanation / Answer

unknowns are : Volume of solution containing 26% NaCl and rest water (74%). Weight of sample leaving the mixture.

Number of independent equatioons that need to be solved are

The densit of 26% NaCl solution relates mass to volume

. The sum of weight fractions of 26 Wt% NaCl solution is equal to1

The sum of weight fractions of 26%wt NaCl solution is equal to 1.

The sum of weight fractions of product solution equal to 1.

The densitty of the 4 wt% solution relates mass to volume.

Mass balance on NaCl.

Mass balance on water

c) Volume of first solution =325 gallons=325*3.78 L (1 gallon= 3.78 L)=1228.5 L, 1000 CC =1L 1228.5 L= 1228.5*1000 cc= 1.2285*106cc

its density = 1.0253 g/cm3, mass of first solution = Volume*density= 1.2285*106*1.0253 gm=1259581 gm

Let the volume of second solution in cc be V2. its mass= V2*1.1944 gm

Total mass entering the mixtute= 1259581+V2*1.1944 gm (1)

NaCl at the outlet is 15% , i.e (1259581+V2*1.1944)*15/100= 1259581*0.15+0.15*1.1944V2=188937.2 +0.179V2 (2)

This is equal to NaCl entering from Stream 1 + NaCl entering from stream 2

=1259581*4/100+V2*1.1944*26/100= 50383.24+0.3105 V2 (3)

As per mass balcne Eq.2 =Eq.3

188937.2+0.179V2= 50383.24+ 0.3105V2

188937.2-50383.24= V2*(0.3105-0.179) , V2= 1053642 ml =1053.642L=278.7 Gallons.

Let the volume of second feed be = V ml

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