The volume of 10 g of gas X is 5.6 litre at NTP. What is the molecular weight of X?
The question asks us to determine the molecular weight of a gas X given its mass and volume at Normal Temperature and Pressure (NTP).
NTP stands for Normal Temperature and Pressure. Under NTP conditions:
A key concept for gases at NTP is the molar volume. One mole of any ideal gas occupies a volume of 22.4 litres at NTP.
We know the volume of the gas at NTP and the molar volume at NTP (22.4 L/mol). We can use the formula:
$$ \text{Number of moles} = \frac{\text{Volume at NTP}}{\text{Molar Volume at NTP}} $$
Substituting the given values:
$$ \text{Number of moles} = \frac{5.6 \text{ L}}{22.4 \text{ L/mol}} $$
Let's perform the division:
$$ \text{Number of moles} = \frac{5.6}{22.4} = \frac{56}{224} = \frac{1}{4} = 0.25 \text{ moles} $$
So, 10 g of gas X is equal to 0.25 moles.
Molecular weight is defined as the mass per mole of a substance. We have the mass (10 g) and the number of moles (0.25 moles).
The formula for molecular weight is:
$$ \text{Molecular Weight} = \frac{\text{Mass}}{\text{Number of moles}} $$
Substituting the values:
$$ \text{Molecular Weight} = \frac{10 \text{ g}}{0.25 \text{ moles}} $$
To simplify the division, we can multiply the numerator and denominator by 4:
$$ \text{Molecular Weight} = \frac{10 \times 4}{0.25 \times 4} = \frac{40}{1} = 40 \text{ g/mol} $$
The molecular weight of gas X is 40 g/mol.
| Quantity | Value | Unit |
|---|---|---|
| Mass of gas X | 10 | g |
| Volume at NTP | 5.6 | L |
| Molar Volume at NTP | 22.4 | L/mol |
| Number of moles | 0.25 | moles |
| Molecular Weight | 40 | g/mol |
The calculated molecular weight of gas X is 40.
| Concept | Definition/Value |
|---|---|
| NTP (Normal Temperature and Pressure) | 0°C (273.15 K) and 1 atm |
| Molar Volume at NTP | 22.4 L/mol (for ideal gases) |
| Number of Moles Formula (from Volume at NTP) | $$ n = \frac{V_{\text{NTP}}}{22.4 \text{ L/mol}} $$ |
| Molecular Weight Formula | $$ M = \frac{\text{mass}}{n} $$ |
The relationship between pressure, volume, temperature, and the number of moles of an ideal gas is described by the Ideal Gas Law:
$$ PV = nRT $$
Where:
At NTP (P = 1 atm, T = 273.15 K), the molar volume is:
$$ V/n = RT/P = (0.0821 \text{ L·atm/(mol·K)}) \times (273.15 \text{ K}) / (1 \text{ atm}) \approx 22.4 \text{ L/mol} $$
This confirms why 22.4 L/mol is used as the molar volume at NTP for calculating the number of moles.
Which of the following correctly represents the number of atoms in one mole of CH3 OH?
A solution of Gallic acid in ethanol was made by dissolving 0.05 mg in 250 mL of ethanol(w/v). This solution in ppm would be ______________.
The molarity of a solution containing 5 gram of Sodium hydroxide (NaOH) in 500 millilitre solution is : (Take atomic weights of elements as : Na = 23, O = 16, H = 1)