How many litres of water should be added to a 30 litre mixture of milk [#permalink]
10 Dec 2020, 07:45
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How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7 : 3 such that the resultant mixture has 40% water in it?A. 5 litresB. 7 litresC. 10 litresD. 11 litres
E. 12 litres
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Re: How many litres of water should be added to a 30 litre mixture of milk [#permalink]
10 Dec 2020, 09:17
Bunuel wrote:
How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7 : 3 such that the resultant mixture has 40% water in it?A. 5 litresB. 7 litresC. 10 litresD. 11 litres
E. 12 litres
Amount of milk originally = (7 /10) * 30 = 21 liters. Therefore amount of water = 9 LitersLet us add x liters of water to make it a 60% Milk and 40% water solutiontherefore \(\frac{21}{9 + x} = \frac{60}{40} = \frac{3}{2}\)21 * 2 = 3(9 + x)42 = 27 + 3x3x = 15x = 5
Option A
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Re: How many litres of water should be added to a 30 litre mixture of milk [#permalink]
13 Dec 2020, 19:19
How to solve this question using scale method ?
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Re: How many litres of water should be added to a 30 litre mixture of milk [#permalink]
13 Dec 2020, 19:37
Bunuel wrote:
How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7 : 3 such that the resultant mixture has 40% water in it?A. 5 litresB. 7 litresC. 10 litresD. 11 litres
E. 12 litres
Another way to do itWhat is constant? The quantity of milk => 30*7/(3+7)=21This 21 now becomes 100-40 or 60%. => 21=x*60/100.......x=21*100/60=35The increase = 35-30=5A _________________
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Re: How many litres of water should be added to a 30 litre mixture of milk [#permalink]
13 Dec 2020, 20:12
IMO AHere the volume of milk remains the same = 30 Litres * 7/10 = 21 LitresNow these 21 litres = 60 % of New TotalVolume Therefore, New Volume = 35And the water in New Volume = 2/5 of 35 = 14
Water Added = 14 - 9 = 5 Litres
Re: How many litres of water should be added to a 30 litre mixture of milk [#permalink]
13 Dec 2020, 20:12
ChemistrySolutionsPercent Concentration
The percentage concentration of any solution is most commonly expressed as mass percent:
Mass % of any component of the solution = (Mass of the component in the solution / Total mass of the solution) x 100
Other methods are:
Volume % of a component =
(Volume of the component/Total volume of the solution) x 100
Mass by volume percentage:
It is the mass of solute dissolved in 100 mL of the solution.
i.e. Mass by Volume percentage =
(Mass of solute in grams/Volume of solution in mL) x 100
Here's a point to be kept in mind :
Whenever we say mass or volume of the solution, you need to add the respective masses and volumes of ALL the components of the solution. Do NOT commit the error of taking the mass or volume of only the solute or solvent in the denominators of the above expressions.
The concentration of a solution is most of the time expressed as the number of moles of solute present in 1 Liter of the solution (also called molarity )
(There are also other ways to express concentration. Please follow this link. )
EXAMPLE:
(a) If 25 moles of NaCl are present in 100 L of a solution wherein H2O is the solvent, then the concentration of the solution is #25/100=0.25 "mol·L"^-1#.
(b) What is the molarity of a solution prepared by dissolving 15.0 g of sodium hydroxide in enough water to make a total of 225 mL of solution?
Solution
Calculate the number of moles of solute present.
Moles of NaOH = 15.0 g NaOH × #(1"mol NaOH")/(40.00"g NaOH")# = 0.375 mol NaOH
Calculate the number of litres of solution present.
Volume = 225 mL × #(1"L")/(1000"mL")# = 0.225 L soln
Divide the number of moles of solute by the number of litres of solution.
Molarity = #(0.375"mol")/(0.225"L")# = 1.67 mol/L
Let's address the question for both percent concentration by mass and for percent concentration by volume.
Percent concentration by mass is defined as the mass of solute divided by the total mass of the solution and multiplied by 100%. So,
#c% = m_(solute)/(m_(solution)) * 100%#, where
#m_(solution) = m_(solvent) + m_(solute)#
There are two ways to change a solution's concentration by mass
Adding more solute - making the solution more concentrated;
Adding more solvent - making the solution more dilute;
Let's take an example to better illustrate this concept. Say we dissolve 10.0g of a substance in 100.0g of water. Our concentration by mass will be
#c% = (10.0g)/(10.0g + 100.0g) * 100% = 9.09%#
Now let's try doubling the mass of the solute; the new concentration will be
However, if we keep the mass of the solute at 10.0g and doubled the mass of the solvent (in this case, water), the concentration will be
#c% = (10.0g)/(10.0g + 2*100.0g) * 100% = 4.76%#
The same is true for percent concentration by volume, which is defined as the volume of the solute divided by the total volume of the solution and multiplied by 100%.
It's easy to see that manipulating either the volume of the solute or the volume of the solvent (or both) would change the solution's percent concentration by volume.
There are two types of percent concentration: percent by mass and percent by volume.
PERCENT BY MASS
Percent by mass (m/m) is the mass of solute divided by the total mass of the solution, multiplied by 100 %.
Percent by mass = #"mass of solute"/"total mass of solution"# × 100 %
Example
What is the percent by mass of a solution that contains 26.5 g of glucose in 500 g of solution?
Solution
Percent by mass =
#"mass of glucose"/"total mass of solution" × 100 % = (26.5"g")/(500"g")# × 100 % = 5.30 %
PERCENT BY VOLUME
Percent by volume (v/v) is the volume of solute divided by the total volume of the solution, multiplied by 100 %.
Percent by volume = #"volume of solute"/"total volume of solution"# × 100 %
Example
How would you prepare 250 mL of 70 % (v/v) of rubbing alcohol
Solution
70 % = #"volume of rubbing alcohol"/"total volume of solution" × 100 %# × 100 %
So
Volume of rubbing alcohol = volume of solution × #"70 %"/"100 %"# = 250 mL × #70/100#
= 175 mL
You would add enough water to 175 mL of rubbing alcohol to make a total of 250 mL of solution.