The solution before dilution and after dilution contains same number of moles, and water is added for dilution.
Option B
<h3><u>Explanation:</u></h3>
Suppose before dilution, the solution contains x moles of KCl in Y liter of water. Now as the concentration got halved, then the solution contains x moles of KCl in 2Y kiters of solution. So the number of moles of KCl in the solution remained constant.
Again, as the solution is diluted to half of the concentration, water must have been added with the solution to make it dilute.
C is the correct answer (CaF2) (sorry dont have subscript)
Explanation: synthesis reaction forms a compound and calcium reacting with fluorine produces Calcium Fluoride (CaF2) chemical name
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m
a) To find the mass after t years:we will use this formula:
A = Ao / 2^n when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)b) Mass after 90 y :by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mgC) Time for 1 mg remaining:when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y