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Answer: The percentage by mass of sulphur in
is 9.36%
Explanation:
Mass percent of an element is the ratio of mass of that element by the total mass expressed in terms of percentage.

Given: mass of sulphur = 32 g/mol
mass of
= 342 g/mol
Putting in the values we get:

The percentage by mass of sulphur in
is 9.36%
Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
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