Answer to Question #130933 in Molecular Physics | Thermodynamics for Johannes Steven

Question

1. The density of water is 1.00 g/ML at 4 degrees Celsius. How many water molecules are present in 2.56 mL at this temperature?

2. For a 4.35 g sample of ammonia gas (NH3), Determine:

(a) Number of ammonia molecules.
(b) Total number of atoms in a sample.
(c) Volume of the gas at room temperature of 25 degrees Celsius and a pressure of 1 atm.

Answer 1

Problem 1:

Density $\rho=\frac{m}{V}$

$\nu=\frac{m}{M}=\frac{N}{N_A}$ whence $N=\frac{m}{M}N_A=\frac{\rho*V}{M}*N_A$

Molar mass of water: $M(H_2O)=18*10^{-3}\frac{kg}{mole}$

Density of water: $\rho=1000\frac{kg}{m^3}$

Volume $V=2.56*10^{[-6}m^3$

Avogadro number: $N_A=6.02*10^{23}mole^{-1}$

So, the number of water molecules:

$N==\frac{10^3*2.56*10^{-6}}{18*10^{-3}}*6.02*10^{23}=0.86*10^{23}$

Problem 2:

a) $\nu=\frac{m}{M}$

Molar mass of ammonia $M=17\frac{g}{mole}$

$\nu=\frac{4.35}{17}=0.26$ moles

b) $\nu=\frac{N}{N_A}$ whence $N=\nu*N_A$

$N=0.26*6.02*10^{23}=1.57*10^{23}$

c) By the Mendeleev-Clapeiron equation:

$PV=\nu*RT$

P is pressure, $P=10^5Pa$

T is absolute temperature (in degrees of Kelvin), $T=t+273=25+273=298$ K

R is universal gas constant $R=8.31\frac{J}{K*mole}$ . Then the volume:

$V=\frac{\nu*RT}{P}$

$V=\frac{0.26*8.31*298}{10^5}=6.4$ (litres)