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A chemistry student is working with solution M, which has a 20% calciu
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12 Jul 2018, 19:17
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A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove threefourths 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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A chemistry student is working with solution M, which has a 20% calciu
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12 Jul 2018, 19:37
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 Answer C



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Re: A chemistry student is working with solution M, which has a 20% calciu
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12 Jul 2018, 19:41
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. \(X*\frac{175}{100} = 70\) \(X*\frac{7}{4} = 70\) \(X = 40\) Option C is the answer
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Re: A chemistry student is working with solution M, which has a 20% calciu
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12 Jul 2018, 19:51
Carlgrc wrote: A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove threefourths 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 threefourths 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=0.3751/4 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
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13 Jul 2018, 17:22
The easiest way to solve this mixture problem is with the general formula (see @VeritasPrepKarishma Mixtures): The formula is: \(\frac{w1}{w2}=\frac{A2A}{AA1}\) 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 \(\frac{1}{4} * \frac{4}{7}=\frac{N37,5}{37,520}\) \(N = \frac{375}{10} + \frac{175}{10} * \frac{1}{7}\) N = 40 Answer C
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Re: A chemistry student is working with solution M, which has a 20% calciu
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13 Jul 2018, 17:57
Carlgrc wrote: A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove threefourths 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 methodVolume 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 11/8=7/8 , that is 7 parts and volume be v. M...................P......N 1...................8.......7 20%..........37.5%...v So \(v=P + \frac{(37.520)}{7}=37.5+\frac{17.5}{7}=37.5+2.5=40%\) C
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A chemistry student is working with solution M, which has a 20% calciu
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13 Jul 2018, 18:20
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 Answer C



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Re: A chemistry student is working with solution M, which has a 20% calciu
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14 Jul 2018, 11:56
Carlgrc wrote: A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove threefourths 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
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30 Jul 2018, 05:17
chetan2u wrote: Carlgrc wrote: A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove threefourths 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 methodVolume 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 11/8=7/8 , that is 7 parts and volume be v. M...................P......N 1...................8.......7 20%..........37.5%...v So \(v=P + \frac{(37.520)}{7}=37.5+\frac{17.5}{7}=37.5+2.5=40%\) 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
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10 Aug 2018, 17:48
dimmak wrote: A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove threefourths 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% Answer: C
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