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
Answer:
Explanation:
Oleic acid originates from an unsaturated fatty acid with 18 carbon atoms and one double bond, which can be found in olive oil and many other vegetable and animal oils and fats. Its economic importance includes the production of soap making and cosmetics etc.
The line-bond structure of the given oleic acid in the question can be found in the attached below.
<em>Same group element have same</em><em><u> Valence electron</u></em><em> and behave similarly in </em><em><u>Chemistry.</u></em>
<u>Explanation:</u>
For example. First group elements Alkali metals:- H, Li, K, Rb, Cs, Fr
Valance electron will take part in forming a bond with other elements and compound will form. All the above-given elements (H-Fr) have valence electron 1 in outer most 'S' shell. All have electronic configuration S1
Behavior: Since valence electrons are the same so the behavior of all the elements in this group is the same. All are metal (from Li-Fr, except Hydrogen), all are very reactive, does not found in native state in the environment, and all react with water.
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
1 mol ---- 40g
X --------- 15g
X = 15/40 = 0,375 moles
:)
