Answer:
3.092 x 10²¹ molecules
Explanation:
Data given:
mass of volatile liquid (m) = 0.23 g
Volume of volatile liquid (V) = 126.4 cc
convert cc to liter
1 cc = 0.001 L
126.4 cc = 126.4 x 0.001 = 0.1264 L
Temperature (T) = 27 °C
Convert °C to kelvin
T(K) = °C + 273
T(K) = 27 + 273 = 300 K
pressure (P) = 760 mm
Convert mm to atm
1 mm = 760 atm
So
P = 1 atm
Solution:
This will be solved in three steps
- First we have to find out molar mass of the liquid
Formula will be used
PV= nRT . . . . . . .(1)
where R is ideal gas constant = 0.08206 L.atm / mol. K
we also know that
n = m/Mr . . . . . . . (2)
where
m is mass
Mr is molar mass
Combine both equation 1 and 2
PV= mRT/Mr . . . . . . . (3)
Rearrange the equation 3
Mr = mRT/PV . . . . . . .(4)
Put values in equation 4
Mr = 0.23 g x 0.08206 L.atm / mol. K x 300 K / 1 atm x 0.1264 L
Mr = 5.658 g.L.atm/mol / 0.1264 atm.L
Mr = 44.8 g/mol
- Now we have to convert mass to moles
Formula used
no. of moles = mass in grams / molar mass
Put values in above formula
no. of moles = 0.23 g / 44.8 g/mol
no. of moles = 0.0051 mol
- Now we have to find no. of molecules
Formula used
no. of molecules = no. of moles x Avogadro's number
where
Avogadro's number = 6.022 x 10²³
put values in above equation
no. of molecules = 0.0051 mol x 6.022 x 10²³
no. of molecules = 3.092 x 10²¹ molecules
0.23 gram contains 3.092 x 10²¹ molecules
