How much water and and how much 32% solution does she need to use?

Question

Can someone anser and explain the following question?
A Chemist dilutes a 32% saline solution with water (0% saline) in order to prouce 150 lters of 20% saline solution. How much water and how much 32%solution does she need to use?

Answer 1

this can be a problem to you if you don't know anything about simultaneous equation or if u don't know how to form simultaneous equation. another thing is u should at least have an idea about chemistry. look

m1v1+m2v2=MV.............1
m1+m2=total volume after dilution=150Lt..............2
now
take m1 is the initial conc of saline soln=32%=0.32
'' v1 volume =?
m2 is the conc of water =0
v2 is the volume of water =?
M is the final conc of the soln = 20%=0.20
V is the final volume of the soln obtained after mixing=150Lt

now
0.32v1+0*v2=0.2*150......................1

v1+v2=150.................2
from eqn 1
0.32v1=30
so v1= 93.75Lt's
so the remaining amount is of water
v2=150-93.75=56.25lts
so you need 56.25litres of water and 93.75litres of 32%saline to prepare the solution you need.

if you have these solution just try to mix at that proportion then u will tell me.

Answer 2

You're trying to get to 20 % solution of saline from mixing 32% saline with water. You are totalling 150 L. . 32% of 150 is 48 liters. 20 % of 150 is 30 liters. Difference is 18 liters. SO, you will need 18 liters of 32% solution and (150 - 18 = 132) 132 Liters of water.

Answer 3

well if the percentage you gave is volumetric go this way

first find the amount of pure saline
pure saline = 0.2 x 150L = 30L
to get 32% saline solution = 30/0.32 = 93.75L
Therefore the amount of water = 150 - 93.75 = 56.25L

Answer 4

Let S be the amount of undiluted saline solution. Then:
.32S=.2(150)
S=93.75 L
Thus she will need 56.25 L of water to dilute with 93.75 L of solution.

Answer 5

x = water, then 150 - x = 32% solution
0(x) + 32(150-x) = 20(150)
Solve this equation and you will have your answers.