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# A chemistry student is working with solution M, which has a 20% calciu

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Manager
Joined: 08 Sep 2017
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A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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12 Jul 2018, 20:17
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51% (02:09) correct 49% (03:00) wrong based on 114 sessions

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A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%

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Intern
Joined: 11 Jul 2018
Posts: 20
A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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12 Jul 2018, 20:37
Assume you have 1000 ml M out of which 200 ml is calcium chloride leading to 20% concentration.
Remove 3/4 of M, which will mean now we have (1000/4) = 250 ml M out of which (200/4) = 50 ml is calcium chloride leading to 20% concentration.
To have 37.5% concentration in 2000 ml we should have 750 ml in 2000 ml.

2000 ml is taken as question says we need twice the volume of solution M in final solution P.
(One small assumption here that twice the volume of P meant twice from original quantity of M and not after removing 3/4th)

In this remove remaining quantities from solution M => 750 - 50 = 700 ml & 2000 - 250 = 1750 ml
These quantities need to come from solution N.

Concentration of calcium chloride in solution N = (700/1750)*100 = 40
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Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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12 Jul 2018, 20:41
3
Consider Solution M ha a volume of 100.
calcium chloride = 20% = 20.

Solution M is reduced by \frac{3}{4}.
New volume of M = 25
calcium chloride = 5

Solution P = N + 25
We know P has double the volume of M.
Therefore, 200 = N + 25
N = 175.

Concentration of calcium chloride required in P = $$37.5%$$ or $$75$$ (200*37.5%)
We already have 5 units of calcium chloride from M, we need 70 more.

$$X*\frac{175}{100} = 70$$
$$X*\frac{7}{4} = 70$$
$$X = 40$$

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Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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12 Jul 2018, 20:51
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Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%

X=Concentration of Ca in N
M=Volume of solution M
N=volume of solution N
P=volume of solution P
Remove three-fourths of M and then add N to create P, P=2M
=> 1/4 M +N=2M=>N=7/4 M
Concentration Of Ca in P=0.375
1/4 M*0.2 + X *7/4M =3/8 *2M=>x=0.4

Anser is C
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A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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13 Jul 2018, 18:22
1
The easiest way to solve this mixture problem is with the general formula (see @VeritasPrepKarishma Mixtures):

The formula is: $$\frac{w1}{w2}=\frac{A2-A}{A-A1}$$

Concentration of Ca in M = 20% = A1
Concentration of Ca in P = 37,5% = A
Concentration of Ca in N =? = A2 = N

If V is the initial volume of M:

Volume of M = w1 = 1/4 V
Volume of P = 2V
Volume of N = w2 --> 2V = 1/4 V + Volume of N --> Volume of N = 7/4 V = w2

$$\frac{1}{4} * \frac{4}{7}=\frac{N-37,5}{37,5-20}$$

$$N = \frac{375}{10} + \frac{175}{10} * \frac{1}{7}$$

N = 40

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Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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13 Jul 2018, 18:57
Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%

weighted average method
Volume of initial solution M is 1/4 and concentration is 20%.But since final volume is 2 times, the volume of remaining M as a part of P is 1/4*1/2=1/8 so 1 part
Volume of solution P is 1 or8 parts and concentration is 37.5%
Volume of N is 1-1/8=7/8 , that is 7 parts and volume be v.

M...................P......N
1...................8.......7
20%..........37.5%...v
So $$v=P + \frac{(37.5-20)}{7}=37.5+\frac{17.5}{7}=37.5+2.5=40%$$

C
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A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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13 Jul 2018, 19:20
Bak 2 chem

M
x volume
.2x cacl
.25x volume
.05x cacl

N
2x-.25x volume
nx cacl

(.05x+nx)/(2x)=.375

n=.7

n/1.75 = .4
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Posts: 1067
Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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14 Jul 2018, 12:56
Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%

let x=calcium chloride concentration of solution N
.2*(1/4)+x*(7/4)=.375*2
x=.4=40%
C
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Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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30 Jul 2018, 06:17
chetan2u wrote:
Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%

weighted average method
Volume of initial solution M is 1/4 and concentration is 20%.But since final volume is 2 times, the volume of remaining M as a part of P is 1/4*1/2=1/8 so 1 part
Volume of solution P is 1 or8 parts and concentration is 37.5%
Volume of N is 1-1/8=7/8 , that is 7 parts and volume be v.

M...................P......N
1...................8.......7
20%..........37.5%...v
So $$v=P + \frac{(37.5-20)}{7}=37.5+\frac{17.5}{7}=37.5+2.5=40%$$

C

Hi Chetan,

It will really help if you can explain the weighted average method and/or maybe paste links pointing to more questions and similar theories. I am struggling to understand, being thick headed!!
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Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

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10 Aug 2018, 18:48
dimmak wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%

We can let amount of solution M = 20 liters, so it has 0.2 x 20 = 4 liters of calcium chloride. We also remove ¾ x 20 = 15 liters of M, which includes ¾ x 4 = 3 liters of calcium chloride. Thus, we have 5 liters of M left (which includes 1 liter of calcium chloride), and we have to add 35 liters of solution N to make 40 liters of solution P (recall that solution P is double the amount of solution M).

We can x = the percent of calcium chloride in solution N and create the equation:

(1 + x/100 * 35)/40 = 0.375

1 + 35x/100 = 15

35x/100 = 14

35x = 1400

x = 40

Alternate Solution:

Assume we start with 4 liters of 20% solution. We will remove ¾ of it, which is 3 liters of 20% solution. Now we add 7 liters of unknown concentration x, and the result is 8 liters of 37.5% solution. Note that we need to end with twice the volume of what we started with, so we end with 8 liters of solution. We can create the following equation:

4(0.20) - 3(0.20) + 7x = 8(0.375)

0.8 - 0.6 + 7x = 3

7x = 2.8

x = 0.40, or 40%

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Re: A chemistry student is working with solution M, which has a 20% calciu &nbs [#permalink] 10 Aug 2018, 18:48
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