Calculate the molality of 0.4mol NaCl dissolved in 500g of water.
[Ar; Na=23, H=1, O=16, Cl=35.5]
Molality is defined as the number of moles of solute per kilogram of solvent.
Given:
- Moles of NaCl = 0.4 mol
- Mass of water = 500 g
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
First, we need to convert the mass of water to kilograms:
Mass of water = 500 g = 500/1000 kg = 0.5 kg
Next, we can calculate the molality using the formula:
Molality = moles of solute / mass of solvent in kg
Molality = 0.4 mol / 0.5 kg = 0.8 mol/kg
Therefore, the molality of 0.4 mol NaCl dissolved in 500 g of water is 0.8 mol/kg.
50cm3 of 0.2M NaOH solution was diuted to 250cm3. What is the concentration of the diluted solution?
To find the concentration of the diluted solution, we can use the equation:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution before dilution
V1 = initial volume of the solution before dilution
C2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
Given:
C1 = 0.2 M (concentration of the initial NaOH solution)
V1 = 50 cm^3 (initial volume of the NaOH solution)
V2 = 250 cm^3 (final volume of the diluted solution)
Plugging in the values to the equation:
C1V1 = C2V2
0.2 M * 50 cm^3 = C2 * 250 cm^3
C2 = (0.2 M * 50 cm^3) / 250 cm^3
C2 = 0.04 M
Therefore, the concentration of the diluted solution is 0.04 M.
A chemist has a 505ml of a 0.125M HCl and wants to dilute it to exactly 0.100M. How much water should be added?
To solve this problem, we can use the equation for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
Given:
C1 = 0.125 M (initial concentration of HCl)
V1 = 505 mL (initial volume of HCl)
C2 = 0.100 M (final concentration desired)
V2 = ? (final volume of diluted solution)
Plugging in the values to the equation:
0.125 M * 505 mL = 0.100 M * V2
V2 = (0.125 M * 505 mL) / 0.100 M
V2 ≈ 632.5 mL
Therefore, you should add approximately 632.5 mL of water to the 505 mL of 0.125 M HCl to get a final concentration of 0.100 M.
Determine the number of moles present in 1.5x10^20 electrons
[L=6.02×10^23]
*
To determine the number of moles present in a certain number of electrons, we need to use Avogadro's constant to convert from the number of electrons to the number of moles.
Given:
Number of electrons = 1.5×10^20 electrons
Avogadro's constant = 6.02×10^23 (mol⁻¹)
Number of moles = Number of electrons / Avogadro's constant
Number of moles = (1.5×10^20 electrons) / (6.02×10^23 mol⁻¹)
Number of moles ≈ 2.49×10⁻⁴ mol
Therefore, there are approximately 2.49×10⁻⁴ moles present in 1.5×10^20 electrons.
The concentration of 14.63g of salt Y in 250cm3 of solution is;
[Y=58.5g/mol]
To find the concentration of the solution, we need to first calculate the number of moles of salt Y using its molar mass and then divide it by the volume of the solution.
Given:
Mass of salt Y = 14.63 g
Molar mass of salt Y = 58.5 g/mol
Volume of solution = 250 cm³
First, let's calculate the number of moles of salt Y:
Number of moles = Mass / Molar mass
Number of moles = 14.63 g / 58.5 g/mol
Number of moles ≈ 0.25 mol
Next, we can calculate the concentration of the solution:
Concentration = Number of moles / Volume of solution
Concentration = 0.25 mol / 0.25 L (since 250 cm³ = 0.25 L)
Concentration = 1 mol/L
Therefore, the concentration of the solution is 1 mol/L.
A gas of mass 0.358g occupies a volume of 400cm3 at stp. What is the relative molecular mass of the gas?
[Vm=22.4dm3mol-1]