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What is the molality of a solution that has 6 mol of CaCl2 in 3 kg of water?
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∙ 7y ago
2 m
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∙ 7y ago
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What is the freezing point of a solution that contains 2.0 mol of CaCl2 in 800.0 g of water Kf for water equals 1.86Cm?
2.0 mol of CaCl2 releases 2.0 mol of Ca+2 ions & 4.0 mol of Cl- ions = 6 moles of ions find molality: 6 moles / 0.800 kg water = 7.5 molal solution dT = kf (molality) dT = 1.86 C (7.5m) dT = 14 Celsius drop in freezing temp the new freezing point is - 14 C
What would you need to do to calculate the molality of 10 mol of NaCl in 200 mol of water?
Convert the 200 mol of water to kilograms of water.
Determine the number of moles of Cl in 1.9 mol CaCl2?
Ca-1(40.08)g=40.8g/mol of Ca Cl-2(35.45)g=70.9g/mol of Cl2 40.08g+70.9g=110.98g/mol CaCl2 110.98g CaCl2/1mol * 1.9mol =210.86 g of CaCl2
A solution containing 14 grams of AgNo3 is added to a solution containing 4.83 g of CaCl2. Find the mass of the precipitate produced.?
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2 Precipitate = AgCl Now find the mol of compound in each solution: 14g AgNO3 x (mol/170g) = .082mol 4.83g CaCl2 x (mol/111g) = .044mol Determine limiting reactant: Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant. Now that you know this you can find the mass of the precipitate .082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left .006mol x (111g/mol) = 0.67g CaCl2
What is the mass of 0.89 mol of CaCl2?
0.89 moles of CaCl2 is equal to 98.8g.
What is the molality of a solution that has 6 mol of CaCl2 in 3km of water?
2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of water.
What is the molality of a solution that has 6 mol of cacl2 in 3kg of water?
2m
What is the freezing point of a solution that contains 2.0 mol of CaCl2 in 800.0 g of water Kf for water equals 1.86Cm?
2.0 mol of CaCl2 releases 2.0 mol of Ca+2 ions & 4.0 mol of Cl- ions = 6 moles of ions find molality: 6 moles / 0.800 kg water = 7.5 molal solution dT = kf (molality) dT = 1.86 C (7.5m) dT = 14 Celsius drop in freezing temp the new freezing point is - 14 C
What is the molality solution that has 3 mol of glucose in 6 kg if water?
This molality is 90,08 g/kg.
What is molality of a solution that has 4 mol of KCI in 0.800 kg of water?
4 mol/0.800 kg
What is the molality of solution that has 3 mol of glucose in 6 Kg of water?
3mol/6kg
What is the molality of a solution that has 3 mol of glucose in 6 kg of water?
3mol/6kg
What is the molarity of a solution that has 6 mol of CaCl2 in 3 kg of water?
The molarity is 2 mol/L.
What is the molality of a solution made by dissolving 2 mole of NaOH in 6kg of water?
0.33 mol/kg
What is the molality of a solution made by dissolving 2 moles of NaOH in 6kg of water?
0.33 mol/kg
What is the molality of a solution that contains 5.10 mol KNO3 in 4.47 kg water?
Molality = moles of solute / kilograms of solventm = 5.10 / 4.47m = 1.14 molal
What is the molality of a solution that has 4 mol of KCl in 0.800 kg of water A.?
4 mol over 0.800 kg
