Please help me!!!?
calculate the amount of strong cetrimide solution (40%w/v) required to make 200ml of cetrimide solution 1%w/v.
Answers:
Gene Guy got it right, more or less.
For 1% (w/v) one needs 2 grams of cetrimide. This is the amount found in 5 ml of the 40% solution.
So the procedure is measure 5 mL of the strong solution and place it in a 200 mL volutmetic flask. Then fill to the "mark" with the appropriate solvent (perhaps water).
don't be ridiculous!
You do it you lazy git
Solve as follows:
(1% sol'n / 40% sol'n) x 200 ml final vol. = 5 ml of 40%
to be added to 195 ml of the appropriate diluent
what does that mean?
I have no idea what you're talking about.
Sorry, I dont understand!!x
STOP CHEATING and WORK IT OUT YOURSELF
ahh well that would go as follows:
45 + 887% / 6776 + (77473 x 774) - 776 / 2 x 2.5566748+78634 = I have no idea!
elemental, Mr. Watson
OK so you start with a 40% (w/v) soln and you want 1% (w/v)
thats a 40 fold dilution (40/1= 40)
total volume is 200ml
so you need 200/40= 5ml
First you have to get the weight (w) in 200 mL of 1% cetrimide solution: 1%=w/v → 0.01=w/200mL.
When you transpose this you will get: w = 2 g.
Then substitute the weight that you got, which is 2 g, to this equation: 40% = w/v → 0.4 = 2g /v, transpose this and you will get v = 5 mL
The amount of 40 % strong cetrimide solution needed is 5 mL.
1% w/v means 1g of solute in 100 ml of solvent (ie 1g/100ml).
Firstly we calculate the amount of solute needed to produce the required cetrimide solution:
Amount of solute needed
= 1% x 200ml
= 0.01g/ml x 200ml
= 2 g
So now we know that the the required solution should contain 2g of cetrimide. We can know proceed to calculate the amount of cetrimide solution (40% w/v) needed to produce the final solution.
Given in the question, strength of cetrimide solution is 40% w/v. Obtained from the previous calculation, amount of cetrimide needed is 2g. We also know that 40% w/v means 40g in a 100ml.
Thus, amount of cetrimide solution 40% needed
= 2g divided by 40%
= 2g divided by 0.40g/ml
= 5 ml #
I am very confident with my answer. If you think 5ml is too little, just look from the question that the original solution is a STRONG cetrimide solution (40%w/v) while the required product is a relatively weak cetrimide solution (1%w/v). Thus very little amount of the original solution is needed to produce the required solution with the same amount of solute (2g). I hope this helps.
Take 5 ml of 40%w/v cetrimide and make the volume upto 200ml using distilled water. That would give you 200ml 1%w/v cetrimide
If the centrimide solution was 100% concentration, you would need 200 x 1/100 = 2 grams
As it is 40% concentration you need
2 x 100/40 = 5 grams
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Answers:
Gene Guy got it right, more or less.
For 1% (w/v) one needs 2 grams of cetrimide. This is the amount found in 5 ml of the 40% solution.
So the procedure is measure 5 mL of the strong solution and place it in a 200 mL volutmetic flask. Then fill to the "mark" with the appropriate solvent (perhaps water).
don't be ridiculous!
You do it you lazy git
Solve as follows:
(1% sol'n / 40% sol'n) x 200 ml final vol. = 5 ml of 40%
to be added to 195 ml of the appropriate diluent
what does that mean?
I have no idea what you're talking about.
Sorry, I dont understand!!x
STOP CHEATING and WORK IT OUT YOURSELF
ahh well that would go as follows:
45 + 887% / 6776 + (77473 x 774) - 776 / 2 x 2.5566748+78634 = I have no idea!
elemental, Mr. Watson
OK so you start with a 40% (w/v) soln and you want 1% (w/v)
thats a 40 fold dilution (40/1= 40)
total volume is 200ml
so you need 200/40= 5ml
First you have to get the weight (w) in 200 mL of 1% cetrimide solution: 1%=w/v → 0.01=w/200mL.
When you transpose this you will get: w = 2 g.
Then substitute the weight that you got, which is 2 g, to this equation: 40% = w/v → 0.4 = 2g /v, transpose this and you will get v = 5 mL
The amount of 40 % strong cetrimide solution needed is 5 mL.
1% w/v means 1g of solute in 100 ml of solvent (ie 1g/100ml).
Firstly we calculate the amount of solute needed to produce the required cetrimide solution:
Amount of solute needed
= 1% x 200ml
= 0.01g/ml x 200ml
= 2 g
So now we know that the the required solution should contain 2g of cetrimide. We can know proceed to calculate the amount of cetrimide solution (40% w/v) needed to produce the final solution.
Given in the question, strength of cetrimide solution is 40% w/v. Obtained from the previous calculation, amount of cetrimide needed is 2g. We also know that 40% w/v means 40g in a 100ml.
Thus, amount of cetrimide solution 40% needed
= 2g divided by 40%
= 2g divided by 0.40g/ml
= 5 ml #
I am very confident with my answer. If you think 5ml is too little, just look from the question that the original solution is a STRONG cetrimide solution (40%w/v) while the required product is a relatively weak cetrimide solution (1%w/v). Thus very little amount of the original solution is needed to produce the required solution with the same amount of solute (2g). I hope this helps.
Take 5 ml of 40%w/v cetrimide and make the volume upto 200ml using distilled water. That would give you 200ml 1%w/v cetrimide
If the centrimide solution was 100% concentration, you would need 200 x 1/100 = 2 grams
As it is 40% concentration you need
2 x 100/40 = 5 grams
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