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To find the molarity (M) of the solution, use the formula:
[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} ]
First, convert 183 mmol to moles:
[ 183 , \text{mmol} = 0.183 , \text{mol} ]
Next, convert 449 ml to liters:
[ 449 , \text{ml} = 0.449 , \text{L} ]
Now, calculate the molarity:
[ \text{M} = \frac{0.183 , \text{mol}}{0.449 , \text{L}} \approx 0.408 , \text{M} ]
Thus, the molarity of the solution is approximately 0.408 M.
What else can I help you with?
What is the term for the concentration expression that relates the moles of solute dissolved in each liter of solution?
Molarity, abbrev. M, units: mol/L, mmol/ml
How many mls of a 100mmolar glycine solution contains .005 moles of glycine?
Some conversion required. (mmolar into mol, or moles into mmol ) Molarity = moles of solute/Liters of solution 100 millimolar = 0.1 M glycine Molarity = moles of solute/Liters of solution manipulate algebraically Liters of solution = moles of solute/Molarity 0.005 mole glycine/0.1 M glycine = 0.05 Liters ( 1000 ml/1 L) = 50 milliliters of solution --------------------------------
What is the molarity of a solution that contains 0.5 mil of calcium acetate per 1.0 L of solution?
To calculate the molarity of a solution, you use the formula: Molarity (M) = moles of solute / liters of solution. In this case, 0.5 mil (which is equivalent to 0.5 mmol or 0.0005 moles) of calcium acetate is present in 1.0 L of solution. Therefore, the molarity is 0.0005 moles / 1.0 L = 0.0005 M.
What is the mass of 10 mmol solution?
The mass of a 10 mmol solution will depend on the molar mass of the solute. To calculate the mass, multiply the number of moles (10 mmol) by the molar mass of the solute in grams/mole.
How many grams are equal to 450 milligrams?
Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters) 5M C6H12O6 = moles C6H12O6/0.450 liters = 2.25 moles C6H12O6 (180.156 grams/1 mole C6H12O6) = 405.351 grams of glucose ( you do significant figures )
What is the term for the concentration expression that relates the moles of solute dissolved in each liter of solution?
Molarity, abbrev. M, units: mol/L, mmol/ml
How many mls of a 100mmolar glycine solution contains .005 moles of glycine?
Some conversion required. (mmolar into mol, or moles into mmol ) Molarity = moles of solute/Liters of solution 100 millimolar = 0.1 M glycine Molarity = moles of solute/Liters of solution manipulate algebraically Liters of solution = moles of solute/Molarity 0.005 mole glycine/0.1 M glycine = 0.05 Liters ( 1000 ml/1 L) = 50 milliliters of solution --------------------------------
What is the molarity of a solution that contains 0.5 mil of calcium acetate per 1.0 L of solution?
To calculate the molarity of a solution, you use the formula: Molarity (M) = moles of solute / liters of solution. In this case, 0.5 mil (which is equivalent to 0.5 mmol or 0.0005 moles) of calcium acetate is present in 1.0 L of solution. Therefore, the molarity is 0.0005 moles / 1.0 L = 0.0005 M.
What is the volume in milliliters of 6.44X10-2 M adenosine triphosphate that contains 1.68 mmol of ATP?
I'm not sure, but I did the problem by taking the mmoles and dividing by molarity to get mL. Since mL and mmol both equal x .001, conversion is not neccesary. 1.68/(6.44 x 10^-2)=26.09 mL like I said, I'm not sure this is right but it makes sense to me
A 25.0 ml of 0.106M KIO3 is put into a 250 ml flask and is titrated to a stoichiometric point with 16.25 ml of a sodium thiosulfate solutiom What Molarity of sodium thiosulfate?
Jodate: IO3- + 6e- + 6H+ --> I- + 3H2Oequivalency to:Thio: 6S2O32- --> 3S4O62- + 6e-25.0 ml * 0.106 mmol/ml (KIO3) = 2.65 mmol KIO3 reacting (1:6) with Thio: 6*2.65 = 15.9 mmol sodium thiosulfate. This is in 16.25 ml, so the molarity is 15.9 mmol / 16.25 == 0.978 M thiosulfate
How do you convert mmol to milliters?
1 mol = 103 mmol Conversely, 1 mmol = 10-3 mol For example: 25 mol x 103 mmol/1 mol = 25000 mmol and, 3.2 mmol x 10-3 mol/1 mmol = 0.0032 mol
How do you calculate the volume in milliliters of a 0.61M sodium nitrate solution that contains 400mmol of sodium nitrate?
By definition, a 0.61M sodium nitrate solution contains 0.61 moles of sodium nitrate per liter, which is equivalent to 0.61 mmol/ml. Therefore, the volume of this solution required to contain 400mmol is 400/0.61 or 6.6 X 102 ml, to the justified number of significant digits.
What is the mass of 10 mmol solution?
The mass of a 10 mmol solution will depend on the molar mass of the solute. To calculate the mass, multiply the number of moles (10 mmol) by the molar mass of the solute in grams/mole.
What is the concentration of a HBr solution in 12.0 mL of the solution is neutralized by 15.0 mL of a 0.25 M KOH solution?
Balanced equation. KOH + HBr -> KBr + H2O everything is one to one, so... Molarity = moles of solute/liters of solution ( change ml to liters ) 0.25 M KOH = moles KOH/0.015 liters = 0.00375 moles of KOH this is as many moles that you have of HBr, so... Molarity of HBr = 0.00375 moles/0.012 liters = a concentration of HBr that is 0.31 M
Is mmol EDTA the same as mmol Ca2?
No, mmol EDTA refers to the amount of EDTA (ethylene diamine tetraacetic acid) present, while mmol Ca2+ refers to the amount of calcium ions present in a solution. They are not the same and represent different chemical species.
How many milliliters of 0.140 M HCl are needed to completely neutralize 41.0 mL of 0.107 M BaOH 2 solution?
To determine the amount of HCl needed to neutralize the Ba(OH)2 solution, we can use the equation: nHCl = nBa(OH)2 (molarity HCl * volume HCl) = (molarity Ba(OH)2 * volume Ba(OH)2) nHCl = (0.107 mol/L * 41.0 mL) / 2 = 2.17 mmol Volume HCl = nHCl / molarity HCl = 2.17 mmol / 0.140 mol/L = 15.5 mL Therefore, 15.5 mL of 0.140 M HCl is needed to neutralize 41.0 mL of 0.107 M Ba(OH)2 solution.
How much sodium bicarbonate would be required to prepare 500ml of a 100mM solution?
500 mL * 100(mMol/mL) = 50 mMol NaHCO3 , hence50 mMol NaHCO3 = 50(mMol) * 84(mg/mMol) = 4200 mg = 4.2 g NaHCO3 in 500 mL
