How many grams of testosterone, C19H28O2 (288.4 g/mol), must be dissolved in 299.0 grams of benzene to reduce the freezing point
1 answer:
Answer: 8.42 grams of testosterone
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
= freezing point constant of benzene=
m= molality
Weight of solvent (benzene)= 299.0 g = 0.299 kg
Molar mass of solute testosterone= 288.4 g/mol
Mass of solute testosterone added = ?
Thus 8.42 grams of testosterone must be dissolved in 299.0 grams of benzene to reduce the freezing point by 0.500°C.
You might be interested in
Answer:
The balanced equation for this reaction will be
→
We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2
So using the unitary method we will get that
- 1 mole of CH4 → 4 mole of 4 mole of fluorine
- 0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction
but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.
- 4 moles of fluorine → 1 mole of CF4
- 0.56 mole →
= 0.14mole of CF4
- 4 moles of fluorine → 4 moles of HF
- 0.56 mole of fluorine → 0.56 mole of HF
now to find the heat released we have the formula as
DELTA H = n * Delta H of product - n *delta H of reactant
where n is the moles of the reactant and product.
note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.