The answer is 3.
Explanation:
It’s the last number and it can’t be 9 because then it would be 48.9 and no 3.
5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.
Explanation:
Data given:
mass of sulphur dioxide = 15.2 grams
conditions is at STP whech means volume = 22.4 litres
atomic mass of sulphur dioxide = 64.06 grams/mole
Number of moles is calculated as:
number of moles = 
Putting the values in the equation:
number of moles = 
= 0.23 moles
Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:
When 1 mole of sulphur dioxide occupies 22.4 litres at STP
Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23
= 5.152 litres is the volume.
According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.
the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

Where
xB = mole fraction of solute=?

p = 22.8 torr

mole fraction is ratio of moles of solute and total moles of solute and solvent
moles of solvent = mass / molar mass = 500 /18 = 27.78 moles
putting the values




mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams
Hoi!!
its true
hope this help!!
:) np