If 1.8 L of water is added to 2.5 L of a 7.0 M KOH solution, what is the molarity of the new solution?
Molarity= Number of moles of solute/Liters of solution 50 grams KOH 700 ML to .7 Liters of h2o Molar Mass of KOH= 56 50 divided by 56 = .89 moles Molarity= .89 mol/.7 L = 1.27 MOLARITY
675
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
M1V1=M2V210mLX0.121 = ? X 17mL=0.0708
1.22Molarity (multiply) 0.076Liters = 0.09272Moles needed to neutralize. 0.09272Moles (divided by) 0.125Liters = 0.74176 Molarity HCl has a Molarity of 0.74176
Molarity= Number of moles of solute/Liters of solution 50 grams KOH 700 ML to .7 Liters of h2o Molar Mass of KOH= 56 50 divided by 56 = .89 moles Molarity= .89 mol/.7 L = 1.27 MOLARITY
The pH of water increase.
The concentration is 1 mol/L or 5,611 g KOH/100 mL solution.
675
What is the molarity of an HCl solution if 43.6 mL of a 0.125 M KOH solution are needed to titrate a 25.0 mL sample of the acid according to the equation below?
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
M1V1=M2V210mLX0.121 = ? X 17mL=0.0708
1.22Molarity (multiply) 0.076Liters = 0.09272Moles needed to neutralize. 0.09272Moles (divided by) 0.125Liters = 0.74176 Molarity HCl has a Molarity of 0.74176
Molarity = moles of solute/Liters of solution ( 220.0 ml = 0.220 Liters ) 0.500 M KOH = moles KOH/0.220 Liters = 0.110 moles KOH (56.108 grams/1 mole KOH) = 6.17 grams solid KOH needed
Molarity = moles of solute/volume of solution 0.500 M KOH = moles KOH/125 ml 62.5 millimoles, or to answer the question precisely, 0.0625 moles KOH
Molarity = moles of solute/Liters of solution get moles KOH 6.31 grams KOH (1 mole KOH/56.108 grams) = 0.11246 moles KOH 0.250 M KOH = 0.11246 moles KOH/XL 0.11246/0.250 = 0.4498 liters = 450 milliliters
Molarity = moles of solute/Liters of solutionLiters of solution = moles of solute/MolarityLiters NaOH = 3.25 moles NaOH/2.5 M NaOH= 1.30 Liters NaOH=============