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USPshnik [31]
3 years ago
11

1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

Chemistry
1 answer:
RSB [31]3 years ago
8 0

Answer : The number of molecules present in nitrogen gas are, 3.48\times 10^{13}

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

PV=nRT

where,

P = Pressure of N_2 gas = 1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm      (1 atm = 760 mmHg)

V = Volume of N_2 gas = 985 mL = 0.982 L    (1 L = 1000 mL)

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 0.0^oC=273+0.0=273K

Now put all the given values in above equation, we get:

(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K

n=5.78\times 10^{-11}mol

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains 6.022\times 10^{23} number of molecules.

As, 1 mole of N_2 gas contains 6.022\times 10^{23} number of molecules

So, 5.78\times 10^{-11} mole of N_2 gas contains (5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13} number of molecules

Therefore, the number of molecules present in nitrogen gas are, 3.48\times 10^{13}

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\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}

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