convert the .2 kg of NaCl to moles of NaCl.
Convert the 10 g of NaCl to moles of NaCl
Molality is moles per kg of solvent. You have 2 kg of solvent, so you now need the moles of NaCl. This is 10 g x 1 mole/58.4 g = 0.171 moles. So, molality = 0.171 moles/2 kg = 0.0856 m = 0.086 m (2 sig figs)
convert the 0.2 kg of NaCl to moles of NaCl - apex
Convert the 10 g of NaCl to moles of NaCl.
10g NaCl = 10/58.5=0.17mol. 2kg=2L of water. 0.17/2=0.085M solution.
The molar mass of NaCl: 58,439 g; the molality is 0,085.
The molar mass of sodium chloride is 58,44 g.
Simply convert the 0.2 kg of NaCl to grams and then to moles, and that would be the answer. Thus, 0.2kg NaCl x 1000g/kg x 1 mole NaCl/58.4 g NaCl =~ 3.42 moles/kg water = 3.42 molal.
You need to know the molar mass of NaCl: 58,44 g.
0,2 kg NaCl is 3,42 mol. The molality is 1,14.
Convert the 200 mol of water to kilograms of water.
Convert the 200 mol of water to kilograms of NaCl
i would opt for the Freezing point. salt decreases the freezing point of water. so if water would normally freeze at 0C, saltwater would freeze at -3C.
Not much! Some of it, a tiny amount, might bond to the water molecules, but as water already has its standard H2O composition, most extra hydrogen will simply bubble out, hydrogen being lighter than water. For details and discussion of hydrogen bonding with water, see Related Links below these advertisements. The solubility of hydrogen gas in water at 0oC is 0.0019 grams of hydrogen per kilogram of water. At 60oC, the solubility is 0.0012 grams of hydrogen per kilogram of water. That is a tiny amount that will dissolve in the water. The rest would simply bubble out as the previous answerer said. Also, most likely, the water would be already saturated with hydrogen since it was in contact with the atmosphere, which contains hydrogen; so unless you took steps to purge the hydrogen from the water to get water not already saturated with hydrogen, all of the added hydrogen would bubble out since the water would be already saturated with hydrogen.
The answer depents on the density of the substance involved. If the substance is water (density = 1) then 3.45 liters would have a mass of 3.45 kilograms. 1 kilogram = 2.2406 lbs so 3.45 kg = 7.61 lbs
Convert the 200 mol of water to kilograms of water.
Convert the 200 mol of water to kilograms of water.
Convert the 200 mol of water to kilograms of water.
58,49 g NaCl---------------------1 N10 g---------------------------------xx= 10/58,49=0,171 N58,49 is the molar mass of NaCl (for the foemula unit).Molality= 0,171/2=0,085
58,49 g NaCl---------------------1 N10 g---------------------------------xx= 10/58,49=0,171 N58,49 is the molar mass of NaCl (for the foemula unit).Molality= 0,171/2=0,085
The number of moles of solute dissolved in 1 L of a solution would be the molarity. As an example, if you had 2 moles of solute in 1 liter the molarity would be 2M.
Molality(m)= moles of solute divided by kilograms of solvent you need to find the moles of solute of chloride first by using a conversion factor. 16.1g of Cl2 = 1mole divided by 70.90g of Cl = 0.23mol then you need to convert 5000g of water which is the solvent to kilograms which to convert it you move the decimal 3 units to the left so it is 5.000kg then you put it into the equation so it would be Molality(m)= 0.23mol divided by 5.000kg of solvent Molality= 0.05
Molality of a solution is defined as the number of gram moles, for molecular compounds, or gram formula unit masses, for ionic compounds, mixed with each kilogram of solvent. Sodium chloride is the solute in the question, and water is the solvent. Avogadro's Number is defined as the number of formula units required to constitute one gram formula unit and has the value of about 6.022 X 1023. Therefore, 10 formula units of NaCl constitutes 10/(6.022 X 1023) or about 1.6606 X 10-23 gram formula unit. The gram molecular mass of water is 18.0518. Therefore, 200 molecules of water contains (200)(18.0518)/(6.022 X 1023) or about 599.5 X 10-23 grams or 0.5995 X 10-23) kilograms. Therefore, the molality of the given mixture is 1.6606/0.59995* or about 2.768. ________ *The number 10-23 appears in both numerator and denominator, and therefore need not be included in this fraction.
mole
at 250C the maximum solubility of sucrose, a common form of sugar, is 200g/100ml water or at a molality of 2. However, a supersaturated solution would be able to hold more sugar.
Kilogram
Exactly 1 kilogram