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To 225 mL of a 0.80M solution of KI a student adds enough water to make 1.0 L of a more dilute KI solution What is the molarity of the new solution?
Wiki User
∙ 12y ago
0.18M
Wiki User
∙ 12y ago
Wiki User
∙ 9y ago.18M
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A solution that does not contain enough solute?
A solution that only contains a small amount of solute, is a dilute or very dilute solution.
What is the molarity of a solution made by dissolving 23.4g of sodium sulphate in enough water to make up a 125ml solution?
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.
What is the molarity of a solution prepared by dissolving 1.56 grams of gaseous HCl in enough water to make 26.8 mL of solution?
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
What is the molarity of 20.2 g of potassium nitrate in enough water to make 250.0 mL of solution?
0.8M
What is the molarity of a solution formed by mixing 20.0 g sulfuric acid with enough water to make 500.0 mL of solution?
0.408 M
What is the molarity of a solution made by dissolving 3.09 moles of Nacl in 1.50 L of solution?
what is the molarity of a solution prepared by dissolving 36.0g of NaOH in enough water to make 1.50 liter of solution?
Calculate the molarity of a solution of glycerol made by dissolving 43.000 mL glycerol at 15 degrees C in enough water to make 200.00 mL of solution?
2.89
How do you determine the molarity of a solution containing 18.2g C6H12O6and enough water to reach 6.3L?
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.
Calculate the molar it’s oh a solution that contains 15.7 of CaCO3 dissolved in enough water to make 275mL of solution?
0.570 M (note the capital M, this is molarity.)
How do you solve freezing point depression problems when given only the molarity?
Calculating the freezing point depression of a solvent requires three pieces of information: 1) the molal freezing point depression of that solvent; 2) the molality of the solution; 3) the number of distinct entities produced in solution per molecular formula unit of solute. Hopefully you meant molality instead of molarity (if the solvent is water and it's a dilute solution, assuming that molality = molarity isn't a bad approximation). I suspect you've been told what 1 and 3 are (or been given enough information to figure them out) but simply didn't mention it; otherwise you're pretty much boned.
2.431 g of h2c2o4 x 2h2o was obtained and dissolved in enough dI water to form 250.0 ml of solution what is the concentration of the resulting of h2c2o4 solution?
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
A solution is made by dissolving 19.52 g of SnF2 into enough water to make 770 mL of solution What is the molarity of the solution?
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.
