I have two that I can not figure out:
#1: calculate the MOLALITY of 10.00 g/L of the salt, if its molar mass is 100.0 g/mol and solution density D=1.100 g/mL
work:
10.00 g NaCl x (1mL/1.100g) x (1kg/1000 mL)= 0.00909 kg
100.0 g x (1 mol NaCl/58.4 g Nacl)= 1.7 mol
1.7/0.0090= 187 MOLALITY
#2 calculate the MOLALITY of 1.00 M solution of the salt, if its molar mass is 100.0 g/mol and solution density D= 1.200 G/mL
work:
(1000*1.000 M)/ (1000*1.2 g/mL - 100 g/mol) = .909 MOLALITY
In order to calculate the molality of a solution, you need to know the amount of solute in moles and the mass of the solvent in kilograms.
For the first question:
1. Start with the given concentration of the salt, which is 10.00 g/L.
2. Convert the density of the solution from grams per milliliter (g/mL) to grams per liter (g/L) by multiplying by 1000. So, 1.100 g/mL * 1000 = 1100 g/L.
3. Calculate the mass of the solvent (water) in kilograms using the given density. Divide the mass of the solvent by 1000. So, 10.00 g * (1 mL / 1.100 g) * (1 kg / 1000 mL) = 0.00909 kg.
4. Find the number of moles of the salt using its molar mass. Divide the given mass of the salt by its molar mass. So, 100.0 g * (1 mol NaCl / 58.4 g NaCl) = 1.7 mol.
5. Finally, calculate the molality by dividing the number of moles of the salt by the mass of the solvent in kilograms. So, 1.7 mol / 0.00909 kg = 187 mol/kg. Therefore, the molality of the solution is 187 mol/kg.
For the second question:
1. Start with the given concentration of the salt, which is 1.00 M.
2. Convert the density of the solution from grams per milliliter (g/mL) to grams per liter (g/L) by multiplying by 1000. So, 1.200 g/mL * 1000 = 1200 g/L.
3. Calculate the mass of the solvent (water) in kilograms using the given density. Divide the mass of the solvent by 1000. So, 1.00 mol * (1000 g / 1 L) * (1 L / 1200 g) = 0.909 kg.
4. Finally, the molality is equal to the given molarity, since the molarity is already in mol/L. Therefore, the molality of the solution is 0.909 mol/kg.