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A 0.205 M solution will have 0.205 moles per liter. 250 milliliters is 1/4 of a liter, so in that sample, there will be 0.05125 moles. NaOH has a molar mass of 39.9971 grams per mole, so 2.05 grams added to 250 mL of water will make the desired solution.
Simple equality will do here as you have three things and require a fourth. (X L)(2.5 M NaOH) = (2.0 L)(0.75 M NaOH) 2.5X = 1.5 X = 0.60 Liters needed ================
Make sure that the equation is balanced. NaOH + HNO3 > > NaNO3 + H2O ( all one to one ) 15.0 grams NaOH (1mol NaOH=39.998g) (1mol HNO3/1mol NaOH) = 0.375 moles of HNO3 Molarity = mols solute/volume solution 2.00M HNO3 = 0.375 mols/X volume = 0.188 Liters or, 188 milliliters.
It is possible only if you evaporate the water.
X/200 * 100 = 2.5 x= 5g
Molarity = moles of solute/Liters of solution ( 918 ml = 0.918 liters )rearranged algebraically,moles of solute = Liters of solution * Molaritymoles of NaOH = (0.918 l)(0.4922 M)= 0.45184 moles NaOH=======================so,0.45184 moles NaOH (39.998 grams/1 mole NaOH)= 18.1 grams sodium hydroxide needed============================
A 0.205 M solution will have 0.205 moles per liter. 250 milliliters is 1/4 of a liter, so in that sample, there will be 0.05125 moles. NaOH has a molar mass of 39.9971 grams per mole, so 2.05 grams added to 250 mL of water will make the desired solution.
Simple equality will do here as you have three things and require a fourth. (X L)(2.5 M NaOH) = (2.0 L)(0.75 M NaOH) 2.5X = 1.5 X = 0.60 Liters needed ================
Make sure that the equation is balanced. NaOH + HNO3 > > NaNO3 + H2O ( all one to one ) 15.0 grams NaOH (1mol NaOH=39.998g) (1mol HNO3/1mol NaOH) = 0.375 moles of HNO3 Molarity = mols solute/volume solution 2.00M HNO3 = 0.375 mols/X volume = 0.188 Liters or, 188 milliliters.
It is possible only if you evaporate the water.
Two parts.Molarity = moles of solute/Liters of solution ( 100 mL = 0.1 Liters )0.5000 M NaOH = X moles/0.1 L= 0.05 moles NaOH--------------------------------now,0.05 moles NaOH (39.998 grams/1 mole NaOH)= 1.99 grams NaOH required====================( could call it 2.00 grams )
X/200 * 100 = 2.5 x= 5g
1280 grams
Calculate the mass (in grams) of sodium sulfide that is needed to make 360ml of a 0.50 mol/L solution
HCl has a molar mass of 36.461 grams per mole. This means that 72.922 grams of HCl are needed per liter of water to make a solution that has a concentration of 2M.
we need 0.8gm NaoH and dissolved in 10 ml of water to make 2N solution of NaoH .
The answer is 7,5g.