0
If 55 ml of a 0.211 m NaHO is diluted to a final volume of 125 ml what is the concentration of naoh in the diluted solution?
Wiki User
∙ 14y ago
Are you sure you mean 0.211 m not 0.211 M Is the concentration molality or molarity? If 55 ml of a 0.211 m NaOH is diluted to a final volume of 125 ml what is the concentration of NaOH in the diluted solution?NaOH not NaHO the compound is Sodium hydroxide Molar mass of NaOH = 23 + 16+ 1 = 40 0.211 moles of NaOH = 0.211 * 40 = 8.44 grams of NaOH per liter of solution 55 ml = 0.055 Liter 8.44 grams of NaOH per liter of solution * 0.055 liter = 0.4642 g of NaOH Moles of NaOH = 8.44 ÷ 40 = 0.211 moles 0.211 moles of NaOH in 125 ml = 0.211 moles ÷ 0.125 L = 1.688 m
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
How many grams of KI are in 25.0ml of a 3.0 (mv) KI solution?
55 ml of a 4.05 M solution of KI solution contains 55*4.05=222.75 millimoles. 20.5 ml of the diluted solution contains 3.8g of KI,so no.of moles of KI=3.8/(mol.wt of KI=165.9) is 22.9 millimoles. molarity of final diluted solution=22.9/20.5=1.117M since the no. of moles of KI present in initial and final solution are same. let.V(in ml) be the final volume of diluted solution. 222.75/V=1.117 V=199.41 ml final volume =199.41 ml
What is the final concentration if 75.0 mL of a 3.50 M glucose solution is diluted to a volume of 400 mL?
A 1 mole glucose solution means that there would be 1 mol/liter. There is only .075 liters, so there is only .075 moles present. .075 moles in 400 milliliters is a .075 to .4 ratio. By expanding the ratio, it is found that there are .1875 moles to 1 liter, making the final concentration 0.1875 M.
2 If 3.86 mL of a standard iron solution 1.00 mL 0.050 mg Fe is diluted to 49.9 mL what is the final iron concentration in mg FemL?
Eventually you are going to have to sit an exam in this - it may be better for you if you ask someone to explain the method. Improved*- Any question that is asked on this website could be on an exam. Wikianswers is suppose to give ANSWERS, not personal opinions. And having any questions asked being on a an exam is none of your concern. If you dont know the answer just say you dont know. Times the original concentration by the original volume, this will give you the amount of iron present in the sample. This amount of iron is diluted (divided) into the final volume yeilding the final concentration. As a quick check you can compare the answer to a common sence ratio of the initial and final volumes and concentrations in this case roughly 4 --> 50ml or 0.2 --> 0.004mg/ml. If your answer is in the same region you can be confident you are correct.
What does equation V1 c1 v2 c2 stand for where c concentration and v is for volume?
v1= initial volume c1= initial concentration v2= final volume c2= final concentration For example, you have 10mL of an unknown substance with a concentration of 0,5mol/L. If you add 50mL, what will the final concentration be. V1= 10mL C1= 0,5mol/L V2= 60mL C2= x 10/0,5=60/x You must start by putting everything in the same mesure. We'll use mL here. So 0,5-->1000mL= 50-->10mL 50x60= 300 300/10= 30 30 is your C2
A student carefully diluted 25.0 ml 0.6 M NaOH solution in 475 ml of distilled water What is the molarity of the diluted solution of base?
Assuming that the 25.0 ml are added to the 475 ml of water, not diluted to that volume:Moles of NaOH in the original 25 ml = 25/1000 x 0.6 = 0.015 moles Final volume = 25 + 475 ml = 500 ml. We have 0.015 moles in 500 ml, so 0.03 moles in a litre, Molarity = 0.03.
If 20.00mL of a 1.50M solution of KBr is diluted to a final volume of 150.0mL what is the new molar concentration?
initial molarity*initial volume= final molarity*final volume Initial molarity= 1.50M Initial volume= 20.00ml Final Volume=150.0ml Thus final molarity =1.50M*20ml/150ml=0.200M. New molar concentration= final molarity
What is the molarity of a solution when 50.0ml of a 4.00M KOH solution is diluted to 200ml. what is the percent concentration of KOH in this final solution. asume the density of 1.0gmole?
The concentration is 1 mol/L or 5,611 g KOH/100 mL solution.
What is the concentration of the hydroxide ion in a solution if 0.12 mol of strontium hydroxide are dissolved in water and the final solution volume is 500 mL?
.48 = [OH-]
How much alcohol is there in 51 percent alcohol?
51% typically means 51% by volume. So if you have a 100ml bottle of 51% alcohol, you have 51ml of pure alcohol inside the bottle, diluted to a final solution volume of 100ml.
How many grams of KI are in 25.0ml of a 3.0 (mv) KI solution?
55 ml of a 4.05 M solution of KI solution contains 55*4.05=222.75 millimoles. 20.5 ml of the diluted solution contains 3.8g of KI,so no.of moles of KI=3.8/(mol.wt of KI=165.9) is 22.9 millimoles. molarity of final diluted solution=22.9/20.5=1.117M since the no. of moles of KI present in initial and final solution are same. let.V(in ml) be the final volume of diluted solution. 222.75/V=1.117 V=199.41 ml final volume =199.41 ml
What is the final concentration if 75.0 mL of a 3.50 M glucose solution is diluted to a volume of 400 mL?
A 1 mole glucose solution means that there would be 1 mol/liter. There is only .075 liters, so there is only .075 moles present. .075 moles in 400 milliliters is a .075 to .4 ratio. By expanding the ratio, it is found that there are .1875 moles to 1 liter, making the final concentration 0.1875 M.
What will be the final volume of a solution prepared by diluting 25 mL of 8.25 M sodium hydroxide to a concentration of 2.40?
For all dilution/ concentration problems you use the simple equation: M1V1 = M2V2 2.40*V2 = 8.25*25 V2 = (8.25*25)/2.40 V2 = 85.9mL Final volume will be 86mL.
2 If 3.86 mL of a standard iron solution 1.00 mL 0.050 mg Fe is diluted to 49.9 mL what is the final iron concentration in mg FemL?
Eventually you are going to have to sit an exam in this - it may be better for you if you ask someone to explain the method. Improved*- Any question that is asked on this website could be on an exam. Wikianswers is suppose to give ANSWERS, not personal opinions. And having any questions asked being on a an exam is none of your concern. If you dont know the answer just say you dont know. Times the original concentration by the original volume, this will give you the amount of iron present in the sample. This amount of iron is diluted (divided) into the final volume yeilding the final concentration. As a quick check you can compare the answer to a common sence ratio of the initial and final volumes and concentrations in this case roughly 4 --> 50ml or 0.2 --> 0.004mg/ml. If your answer is in the same region you can be confident you are correct.
What is the concentration in mv percent of a solution prepared from 50 g NaCl and 2.5 L of water?
Chemists often express the concentration of an unsaturated solution as the mass of solute dissolved per volume of the solution. This type of concentration is expressed as percent relationship. The mass/volume percent is also referred to as the percent (m/v)The solubility of NaCl at room temperature is 36g/100mL. Clearly in the problem stated above, the NaCl-H2O solution is quite dilute and much below the saturation point.The following formula is useful in determining the mass/volume percent:m/v% = [ mass solute(g) / volume of solution (mL) ] x 100%2.5 L can be expressed in mL.2.5L = 2500mLtherefore the m/v% = (50g NaCl / 2500 mL H2O) * 100%m/v% = 0.02 * 100%m/v% = 2%Therefore the concentration of this solution as a m/v percent is 2%.Note1: The question is a bit misleading since it never states what is the desired final volume of the NaCl-H2O solution. The required number of grams of solute should never be added directly to the desired final volume of solvent, because volume generally changes when solutions are made.When preparing solutions, the correct procedure is to dissolve the solute in a volume of solvent smaller than the final desired volume. The resulting solution is then diluted to the desired final volume by addition of more solvent. See note 2 for preparation.Note2: To actually prepare this solution, it would be ideal to have the appropriate volumetric glassware. Lets say all you had was a 2L volumetric flask, a 500mL volumetric flask and a 3L storage container.What you could do is add the 50g of NaCl to the empty 2L volumetric flask, then fill the flask with water until you reach the 2L calibration mark. You could then transfer that solution to the 3L storage container.You would then fill the 500mL volumetric flask with water to its calibration mark, and then transfer the water to the storage container. The result would be a 2% (m/v) saline solution with a final volume of 2.5L contained in a 3L storage container.
What is the effect on the molarity of a solution of adding more solvent to a solution?
Adding more solvent to a solution decreases the molarity of the solution. This is based on the principle that initial volume times initial molarity must be equivalent to final volume times final molarity.
What does equation V1 c1 v2 c2 stand for where c concentration and v is for volume?
v1= initial volume c1= initial concentration v2= final volume c2= final concentration For example, you have 10mL of an unknown substance with a concentration of 0,5mol/L. If you add 50mL, what will the final concentration be. V1= 10mL C1= 0,5mol/L V2= 60mL C2= x 10/0,5=60/x You must start by putting everything in the same mesure. We'll use mL here. So 0,5-->1000mL= 50-->10mL 50x60= 300 300/10= 30 30 is your C2
A student carefully diluted 25.0 ml 0.6 M NaOH solution in 475 ml of distilled water What is the molarity of the diluted solution of base?
Assuming that the 25.0 ml are added to the 475 ml of water, not diluted to that volume:Moles of NaOH in the original 25 ml = 25/1000 x 0.6 = 0.015 moles Final volume = 25 + 475 ml = 500 ml. We have 0.015 moles in 500 ml, so 0.03 moles in a litre, Molarity = 0.03.
