The nonvolatile, nonelectrolyte TNT (trinitrotoluene), C7H5N3O6 (227.1 g/mol), is soluble in benzene C6H6. How many grams of TNT
are needed to generate an osmotic pressure of 5.14 atm when dissolved in 279 ml of a benzene solution at 298 K. grams TNT
1 answer:
Answer:
Explanation:
formula of osmotic pressure is as follows
p= n RT
n is mole of solute per unit volume
If m be the grams of solute needed
m gram = m / 227.1 moles
m / 227.1 moles dissolved in .279 litres
n = m / (227.1 x .279 )
= m / 63.36
substituting the values in the osmotic pressure formula
5.14 = (m / 63.36) x .082 x 298
m / 63.36 = .21
m = 13.32 grams .
You might be interested in
Answer:
http://hyperphysics.phy-astr.gsu.edu/hbase/Biology/imgbio/treecycle.p ng
Explanation:
Answer:
E for sphere
Explanation:
A sphere is like a ball
A table decreased in price by
3
5
. After the reduction it was priced at £40. What was the original price of the table?
Answer:
mole fraction of Oxygen = mole of oxygen/ total mole
Explanation:
