(0.23)(1.04g/mL)(1000mL/L)(1 mol C2H6O2/62g C2H6O2) = 3.86 M
(.20)(1.03g/mL)=.260
.260/(molar mass of ethylene glycol)= .260/62.07g/mol=3.32Molarity
Molarity is the no of moles of solute dissolved per litre of a solution. now if u want to find it from the percentage purity , here is the formula for that Molarity = % purity x density x 10 ___________________ molar weight of the solute note : density is usually given %purity problems, if its not u can evualvate it from from formula { d= mass/volume} i hope it solves the problem
A: 10.172 M HCl. Percents of acids like this one is normally given in % w/w, or percent weight by weight. So we need to find the mass of the solution. Molarity is moles of solute per liter solution, so we will calculate this for 1L of solution. The density of 32% HCl is 1.159 g/mL. So assuming we have one liter or 1000 mL, the weight of the solution is 1159 g. If 32% of this is HCl, we have 1159 g solution * 0.32 g HCl/g solution = 370.88 g HCl. The molar mass of HCl is 36.461 g/mol, so 370.88 g HCl * 1 mol HCl/36.461 g HCl = 10.172 mol HCl in 1 L of solution. 10.172 mol HCl/1L solution = 10.172 M HCl.
Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!). Example: What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution? Solution: 11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2 100 mL x 1 L / 1000 mL = 0.10 L molarity = 0.10 mol / 0.10 L molarity = 1.0 M http://chemistry.about.com/od/lecturenotesl3/a/concentration.htm http://www.tpub.com/content/MIL-SPEC/MIL-P/MIL-P-71158/MIL-P-7115800013.htm http://www.tpub.com/content/armymedical/md0837/md08370139.htm
It's very easy as the molarity and normality of NaOH is the same. so all you need to do is to decide how much (volume) of that you need. the molar mass of NaOH is 40.5 gr, so 5 molar (normal) solution needs times five of that in 1 litre. 5*40.5=202.5 dissolve it in 1 liter water (be careful as the dissolving process is highly exothermic) Now you have 1 litre of 5 normality NaOH.
5.84 M
6M
molarity of 5% NaCl solution would be 1.25M.
The concentration is 1 mol/L or 5,611 g KOH/100 mL solution.
Molarity is the no of moles of solute dissolved per litre of a solution. now if u want to find it from the percentage purity , here is the formula for that Molarity = % purity x density x 10 ___________________ molar weight of the solute note : density is usually given %purity problems, if its not u can evualvate it from from formula { d= mass/volume} i hope it solves the problem
17 M
approximalety 2.94 molar
Is the makeup of the solution expressed as "percent by mass"? If so, to calculate molarity (or normality), you have to also know the density of the solution Step 1. Lets say the solution is 14%, and the density is 1.09 g/mL. We can write the following: (14 grams solute/100 grams solution) (1.09 grams solution/ mL solution) Step 2. Multiplying and cancelling from step 1 gives you 15.26 grams solute / 100 mL solution. Multiplying top and bottom by 10 gives you 152.6 grams solute per liter. Step 3. Molarity is number of moles per liter. Divide the 152.6 grams of the solute by the forumua weight (or molecular weight) of the solute, and you have the number of moles of solute. This number is therefore the molarity of the solution. If the solution is "percent by volume", the number you have is number of grams per 100 mL. Multiply by 10, and you have grams per liter. Then divide by the formula weight, and you have the molarity.
If the density is 1.0 g/ml, one liter of the solution will weigh 1000 grams. 3.0 % of this mass or 30 grams of it is constituted of H2O2. The molar mass of H2O2 is 2 (1.008 + 15.999) = 34.014. The molarity of this solution is therefore 30/34.014 = 0.88, to the justified number of significant digits.
Mass percent = grams of solute/total grams of solution Mole fraction = mols component/total mols mix. Molarity = mols solute/L solution Molality = mols solute/kg solvent Hope this helps :)
The density of sodium chloride is 2,165 g/cm3.
This density is 1,0707 g/cm3 at 20 0C.
50% HBF4 in water. MW of HBF4 is 87.81 g density 1.410 g/mL. 1 L solution contains = (1.410 x 1000) x 50% g = 705 g = 705/ 87.81 moles = 8.029 moles So, the solution is 8.029 M.