The volume of mixture of 2gms of helium and 7gms of nitrogen under STP conditions of

A) 22.4lit B) 11.2lit
C) 16.8 D) 5.6lit

2g*(1 mole/4.003g)=0.5 moles of He

7g*(1 mole/14.01g)=0.5 moles of N

Together we have a total of 1 mole of gas.

At STP, 1 mole of a gas has a volume of 22.4L. That is just a relationship that you must commit to memorization.

Options (c)

To find the volume of the mixture of helium and nitrogen under STP conditions, we first need to determine the moles of each gas.

Step 1: Calculate the moles of helium (He)
To find the moles of helium, we use the formula:
moles = mass / molar mass

The molar mass of helium (He) is 4 g/mol.
So, moles of helium = 2 g / 4 g/mol = 0.5 mol

Step 2: Calculate the moles of nitrogen (N2)
To find the moles of nitrogen, we use the same formula:
moles = mass / molar mass

The molar mass of nitrogen (N2) is 28 g/mol.
So, moles of nitrogen = 7 g / 28 g/mol = 0.25 mol

Step 3: Add the moles of helium and nitrogen to find the total moles of the mixture.
Total moles of the mixture = moles of helium + moles of nitrogen
Total moles = 0.5 mol + 0.25 mol = 0.75 mol

Step 4: Use the ideal gas law to find the volume of the mixture
The ideal gas law states that:
PV = nRT

Where:
P = pressure (which is 1 atm for STP)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L*atm/(mol*K))
T = temperature (which is 273 K for STP)

Rearranging the ideal gas law equation, we get:
V = (nRT) / P

Substituting the values in, we get:
V = (0.75 mol * 0.0821 L*atm/(mol*K) * 273 K) / 1 atm
V = 15.89025 L

Since the volume should be given with one decimal place, the volume of the mixture of helium and nitrogen under STP conditions is approximately 15.9 L.

Therefore, none of the given options (A) 22.4 L, (B) 11.2 L, (C) 16.8 L, or (D) 5.6 L are correct.