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The answer is 0,505.
m = moles/kg = 3.00/1.50 = 2.00 molal
You need to know the molar mass of NaCl. Then 10g/molar mass = moles.
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.
Percent concentration could mean many things...MOLARITY is defined as the moles of solute per unit volume of solution so 5 moles of NaCl in one liter of solution would be 5M (molar) NaCl solution.(Note: Solution is the solvent and solute combined, usually the solid is added and then the solution is filled to a certain line once the solid has dissolved.)MOLALITY is defined as the moles of solute per kilogram of solvent (NOT solution). So 5 moles of NaCl in one kilogram of water makes 5 mol/kg NaCl. (sometimes the symbol m is used for molality but is often confused with the unit of meters).For other types of "percent concentration" of solutions check:http://en.wikipedia.org/wiki/ConcentrationA cursory glance seemed to confirm the accuracy of the information.
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
Convert the 200 mol of water to kilograms of water.
The concentration of NaCl is 263 g/L
convert the .2 kg of NaCl to moles of NaCl.
Molality(m)= moles of solute divided by kilograms(kg) of solvent you need to find the moles of NaCl by using a conversion factor 70g NaCl = 1mol divided by 58.44g NaCl =1.20mol then you need to convert 300g of water which is the solvent to kilograms by moving the decimal over 3 units to the left which makes .300 kg of solvent Molality= 1.20mol divided by .300kg Molality=4.00
2.79
0.630 g NaCl x (1 mol NaCl/58.44 g NaCl) = 0.0108 mol NaCl525 g H2O x (1 kg/1000g) = 0.525 kg H2O0.0108 mol NaCl/0.525 kg H2O = 0.0206 molality
The answer is 0,505.
2mol/kg