0.23 gram of volatile liquid occupies 126. 4cc a 27 degree celsius and 760 mm pressure calculate the molecular weight of liquid how many molecules of liquid are present in 0.23 gram?

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

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

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

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

kdkdkrnmcnc

explanation:

wisjndnfnnjjdjdjxn

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.