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# What is the molarity of a solution that has 6 mol of CaCl2 in 3 kg of water?

The molarity is 2 mol/L.

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## Related Questions

2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of water.

molarity is #moles divide by # liters, so 3.0 divided by 0.500 is 6.0 molarity (2 siginficant figures is all you are allowed)

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

Molarity=no. of moles of solute/vol of solution in(L) molarity=0.4/9.79=0.04086 mol L-1=0.04086 M.

Molarity = moles of solute/Liters of solution Molarity = 6 Moles NaCl/2 Liters = 3 M NaCl ========

The moLaRity of this solution ie 2 mol/L or 2M KI (it is a 'two molair potassium iodide' solution)

You'll need: 0.450 (L) * 25 (mol/L) = 11.25 mol CaCl2 and add water to it up to 450 mL.To be weighted:11.25 (mol CaCl2) * [40.08 + 2*35.45](g/mol CaCl2) = 1491.736 g = 1500 g =1.5 kg CaCl2 (CaCl2 as dry substance!, not hydrated)

95g NaOH * 1 mol NaOH/ = 2.375 mol NaOH 40g NaOH Molarity (M) = mol / L M = (2.375) / (0.450) M = 5.28

Molarity (M) = mol solute /L solvent mol solute = 3 mol glucose L solvent = 6kg water = 6L water 3mol/ 6L = .5M

Molarity= moles/Liters To change grams to moles you divide by the mole weight which is listed on the periodic table. Mol= grams/mol weight The Mole weight of Magnesium Chloride is 59.8 grams/mol Mol=128g/59.8 Mol=2.14 Now, you put the number of moles and Liters into the equation Molarity=2.14 mol/1L Molarity=2.14 So, the molarity is 2.14 M

Molarity=Moles of solute/Liters of solution1.350 g KOH = 0.02406 mol KOH0.02406 mol/0.2500 L =0.09624 M of KOH

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