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]
To find the relative molecular mass of the gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, the pressure is 1 atm)
V = volume (400 cm³ = 0.4 dm³)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (at STP, the temperature is 273 K)
Rearranging the equation to solve for n:
n = PV / RT
Given:
P = 1 atm
V = 0.4 dm³
R = 0.0821 L·atm/(mol·K)
T = 273 K
Substituting the given values into the equation:
n = (1 atm * 0.4 dm³) / (0.0821 L·atm/(mol·K) * 273 K)
n ≈ 0.0186 mol
Next, we can calculate the molar mass of the gas using the mass and number of moles:
Molar mass = mass / number of moles
Given:
Mass = 0.358 g
Number of moles = 0.0186 mol
Molar mass = 0.358 g / 0.0186 mol
Molar mass ≈ 19.3 g/mol
Therefore, the relative molecular mass of the gas is approximately 19.3 g/mol.
How much water should be added to 30cm3 of 0.2M NaCl solution to obtain 0.02M solution?
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 = 0.2 M (initial concentration of NaCl)
V1 = 30 cm³ (initial volume of NaCl solution)
C2 = 0.02 M (final concentration desired)
V2 = ? (final volume of diluted solution)
Plugging in the values to the equation:
(0.2 M)(30 cm³) = (0.02 M)(V2)
V2 = (0.2 M)(30 cm³) / 0.02 M
V2 = 300 cm³
Therefore, to obtain a 0.02 M NaCl solution, you should add 300 cm³ of water to the 30 cm³ of 0.2 M NaCl solution.
A concentration of 500ppm is the same concentration as
To determine the concentration in a different unit, we need to convert parts per million (ppm) to the desired unit.
1 ppm is equal to 1 part per million, which means that there is 1 unit of a substance for every 1 million units of the total solution.
To convert ppm to a concentration with a different unit, we need to know the conversion factor or the definition of the desired unit.
For example, if we want to convert ppm to a concentration in milligrams per liter (mg/L), we would need to know the density of the solution. However, without additional information, we cannot determine the specific concentration unit that is equivalent to 500 ppm.
8. A concentration of 500ppm is the same concentration as
To convert parts per million (ppm) to a concentration in another unit, we need to know the specific conversion factor for that unit. Without additional information, we cannot determine the specific concentration unit that is equivalent to 500 ppm.
How many moles of calcium ions are in 40cm3 of 0.45M calcium chloride solution?
To find the number of moles of calcium ions in a solution, we need to use the molarity and volume of the solution and the stoichiometry of the compound.
Given:
Volume of solution (V) = 40 cm³
Molarity of calcium chloride solution (M) = 0.45 M
We know that calcium chloride (CaCl2) dissociates into one calcium ion (Ca2+) and two chloride ions (Cl-). So, the molarity of calcium ion will be the same as the molarity of the calcium chloride solution.
Number of moles = Molarity * Volume
Number of moles of calcium ions = 0.45 mol/L * 40 cm³
Since volume is in cm³ and molarity is in mol/L, we need to convert cm³ to liters.
1 L = 1000 cm³
1 cm³ = 0.001 L
Number of moles of calcium ions = 0.45 mol/L * (40 cm³ * 0.001 L/cm³)
Number of moles of calcium ions = 0.45 mol/L * 0.04 L
Number of moles of calcium ions = 0.018 moles
Therefore, there are 0.018 moles of calcium ions in 40 cm³ of 0.45 M calcium chloride solution.
One mole of ammonia gas, NH3
I. contains 6.02x10^23 molecules
II. the same number of molecules as in 2g of hydrogen gas.
III. contains 3mol of hydrogen atoms and 1 mol of nitrogen atom
IV. has mass of 22.4g
The correct statements are:
I. One mole of ammonia gas, NH3, contains 6.02x10^23 molecules. This is derived from Avogadro's constant, which states that one mole of any substance contains 6.02x10^23 particles, whether they be atoms, molecules, or ions.
III. One mole of ammonia gas, NH3, contains 3 moles of hydrogen atoms and 1 mole of nitrogen atom. This is based on the balanced chemical formula of ammonia, which consists of one nitrogen atom and three hydrogen atoms.
IV. One mole of ammonia gas, NH3, has a molar mass of 17 g/mol. The molar mass of nitrogen (N) is 14 g/mol, and the molar mass of hydrogen (H) is 1 g/mol. Adding these together (14 + 3) gives a molar mass of 17 g/mol. However, the statement regarding 22.4 g is not accurate and does not apply to ammonia gas. The molar mass of ammonia is not 22.4 g/mol.
Therefore, the correct statements are I, III, and IV.
2.8dm3 of oxygen gas at stp contains the following number of molecules
[ Vm =22.4dm3/mol, L=6.02×10^23 ]
To determine the number of molecules in 2.8 dm³ of oxygen gas at STP (Standard Temperature and Pressure), we need to use Avogadro's constant and the molar volume at STP.
Given:
Volume of oxygen gas (V) = 2.8 dm³
Molar volume (Vm) at STP = 22.4 dm³/mol
Avogadro's constant (L) = 6.02×10^23 mol⁻¹
First, we need to calculate the number of moles of oxygen gas using the molar volume:
Number of moles = Volume / Molar volume
Number of moles = 2.8 dm³ / 22.4 dm³/mol
Number of moles = 0.125 mol
Next, we can calculate the number of molecules using Avogadro's constant:
Number of molecules = Number of moles * Avogadro's constant
Number of molecules = 0.125 mol * 6.02×10^23 mol⁻¹
Number of molecules = 7.525×10^22 molecules
Therefore, 2.8 dm³ of oxygen gas at STP contains approximately 7.525×10^22 molecules.