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forsale [732]
3 years ago
7

By mistake, you added salt instead of sugar to the oil. How can you remove the salt?

Chemistry
2 answers:
shutvik [7]3 years ago
8 0

Answer,

Add water to dissolve the salt solution.Shake the solution,oil and water are immiscible liquids therefore oil will not dissolve.Put the solution in a separating funnel and leave the solution undisturbed for a few minutes.Water is denser than oil there oil will come to the top.Power the lower layer of water in to a beaker.Boil the water until water turns into vapour leaving the salt behind.

aivan3 [116]3 years ago
4 0

Salt is ionic, and oil is covalent. If you add water, the salt will dissolve in the water, and the oil will float to the top. You could then skim the oil and start over with sugar. To double-check, you could even make sure the oil does not conduct electricity!

Explanation:

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Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissol
Stells [14]

Here is the complete question.

Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol (CH3CH2OH; density = 0.7893 g/mol). Round to four significant digits

Answer:

0.9567 mol

Explanation:

Given that:

mass of glycerol = 1.61 g

molar mass of glycerol = 92.1 g/mol

no of mole = \frac{mass}{molar mass}

∴ number of moles of glycerol (n_{glycerol}) = \frac{1.61}{92.1}

= 0.0175 mol

Volume of ethanol = 22.60 mL

Density of ethanol = 0.7893 g/mL

Since Density = \frac{mass}{volume}

∴  mass of ethanol = density of ethanol × volume of ethanol

mass of ethanol =  0.7893 g/mL × 22.60 mL

mass of ethanol =  17.838 g

Number of moles of ethanol (n_{ethanol}) = \frac{17.838}{46.0684}

= 0.387 mole

∴ the mole fraction of the solvent can be determined as:

X_{solvent} = X_{ethanol} = \frac{n_{ethanol}}{n_{glycerol}+n_{ethanol}}

=\frac{0.387 mol}{(0.0175+0.387)mol}

= \frac{0.387mol}{0.4045mol}

= 0.95673671199

≅ 0.9567 mol

∴ The mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol is = 0.9567 mol

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