[ 100 (gNaOH) / 40.0 (g/molNaOH) ] in 1.65 L is 1.151 M = 1.515 mol/L
what is the molarity of a solution prepared by dissolving 36.0g of NaOH in enough water to make 1.50 liter of solution?
The molarity of KNO3 is 0,058.
The molarity of a solution made by dissolving 23,4 g of sodium sulphate in enough water to make up a 125 mL solution is 1,318.
Molarity=Moles of solute/Liters of solution1.350 g KOH = 0.02406 mol KOH0.02406 mol/0.2500 L =0.09624 M of KOH
Find moles of HCl first. 1.56 grams HCl (1mole HCl/36.458 grams) = 0.0428 moles HCl Molarity = moles of solute/volume of solution Molarity = 0.0428 moles/26.8 ml = 0.00160 milli-Molarity, or more to the point; = 1.60 X 10^-6 Molarity of HCl
Get moles sodium sulfate first.15.5 grams Na2SO4 (1 mole Na2SO4/142.05 grams) = 0.1091 moles Na2SO4Now,Molarity = moles of solute/ Liters of solution ( 35 ml = 0.035 Liters )Molarity = 0.1091 moles Na2SO4/0.035 Liters= 3.1 M Na2SO4 solution==================
molarity = no. of moles of solute/liter of solution no. of moles of I2 = mass in grams/molar mass = 4.65/253.81 = 0.01832 mol M = 0.01832 mol/0.235 L = 0.0780 mol/L
Molarity = moles solute/liter solutionmoles solute = 7 g NaCl x 1 mol NaCl/58 g NaCl = 0.12 moles NaClliters of solution = 450.0 ml x 1 L/1000 ml = 0.450 litersmolarity = 0.12 moles/0.450 liters = 0.268 M = 0.3 M (one sig fig)
the molarity is found by: 19.52g/770mL x 1 mole/156.7g x 1000mL/1L=0.618 mole i got 156.7g by using the chemical makeup of the solution SnF2: Sn's atomic number being 118.7 and rounding F's atomic number to 19.
Divide grams (mass) by molar mass to find moles58.44 (g NaCl/L) / [22.99+35.45](g NaCl/mol NaCl)= 1.000 mol/L NaCl
Look up or calculate the molecular weight/molar mass of FeCl3. Then... 40.0 g FeCl3/MW FeCl3/0.275 L = M (concentration of FeCl3 in solution)
Molarity = moles of solute / total volume of solution (L or dm3) Moles LiBr = 97.7 g / (6.94 + 79.9) g mol-1 =1.13 mol M = n / V = 1.13 mol / 0.7500 L = 1.50 M There are only three significant figures in the mass of the solute but four in the volume, therefore, answer can not be more than three significant figures.
The question, as worded, is a little ambiguous. Rather, the question you should be asking is “What is the molarity of a 125 ml aqueous solution containing 10.0g of acetone?” Acetone is roughly 58 grams per mole. Therefore, a 125 mil solution with 10 g of acetone would contain roughly 0.17 moles, and the molarity would be roughly 1.4See the Related Questions for more information about how to calculate the molarity of a solution
Moles of solute ---------------- Liters of solution So, convert your grams to moles by dividing by the molar mass (add together the atomic masses of the solution) and then that will give you your moles. Divide that number by your 6.3 Liters. That number is your molarity.
Dissolving takes place when you are trying to make a solution out a mechanical mixture by adding a solute (something that gets dissolved) and a solvent (something that does the dissolving) together and when there is enough solvent to dissolve the all of the solute. When all of the solute is dissolved it becomes a solution which usually appears to be clear.
Iron (III) chloride has the molecular formula of FeCl3. Its molecular weight is 162.2 grams per mole. Concentration is moles of solute divided by volume of solution. Therefore, the answer is .224 moles per liter.
Moles of glucose = 80g/180g/mol = 0.444 moleMolarity = 0.444/0.75L = 0.5926 M
to make the solution basic enough so that NH be prepared for Fmoc protection
Molarity (concentration ) = moles of solute/Liters of solution 250.0 ml = 0.250 liters 2.431 grams H2C2O4 * 2H2O ( 1mole cpd/ 126.068 grams) = 0.01928 moles H2C2O4 * 2H2O Molarity = 0.01928 moles cpd/0.250 liters = 0.07712 Molarity
2.0 mol NaCl / 2.0 Litre = 2.0/2.0 = 1.0 mol/L = 1.0M NaCl
The molarity is 0.001255. Should you really be asking an AP Chem question on Wiki Answers, anyways?
4.0 (g/L) / 40 (g/mol NaOH) = 0.10 (mol/L NaOH) = 0.10 M
[117(g NaCl) / 58.5(g NaCl/mol NaCl)] / 40.0(L solution) = [117/58.5]/40.0 = 2.00(mol NaCl) / 40.0(L) = 0.0500 mol NaCl / L solution = 0.0500 M
80 gNaOH / 500 mL = (80 gNaOH / 40 gNaOH/molNaOH) / 0.500 L = 2.0 mol / 0.500 L = 4.0 mol/L