Molality = moles of solute / kg of solvent.

1 ppm is equivalent to 1 part out of 1 million (10⁶ ) parts.
∴Mass percent of 15 ppm chloroform in water  

⇒1.5 x 10^-3 g chloroform present in 100 g water
Thus, 15g chloroform will be present in water  

Molecular mass of glycerine = 92 g
Molecular mass of water  = 18 g

No. of moles of glycerine = 46/92  = 0.5 moles
No. of moles of water = 36/18 = 2 moles

Thus, Mole fraction of  glycerine = No. of moles of glycerine/(No. of moles of glycerine + No. of moles of water) = 0.5/(2 + 0.5) = 0.20

12.5% w/w means 12.5 g in 100 g of solution.

Weight of solvent = 100 g –12.5 g = 87.5 g.
Number of moles of sulphuric acid = 12.5/ 98 = 0.127 mol

⇒ molality = 0.127 x 1000/87.5 = 1.45 m

When both the solute and solvent are in solid state, then the solution is called a solid solution or solid sol .

Solid Sol includes  gems, photochromic glass, alloys, ruby stone, coloured glass, vision glasses , etc.

In 3M there will be 3 moles per litre therefore in 3 litres there will be  9 moles.

We use  parts per million  to express the concentrations of  solutions  that contain very, very small amounts, often called  trace amounts , of a given  solute.

More specifically, a solution 's concentration in  parts per millions  tells you the number of parts of solute present  for every

10⁶   = 1,000,000

parts of solution . You can thus say that a  1 ppm  solution will contain exactly  1 g  of solute  for every  10⁶ g  of solution.

In your case, you know that you have

in exactly
300.0 g = 3.000 ⋅10² g solution

This means that you can use this known composition as a conversion factor to  scale up  the mass of the solution to  10⁶ g

Since this represents the mass of lead present in exactly  10⁶ g  of solution, you can say that the solution has a concentration of

The molecular weight of NaCl is 58.44 g/mol

So, one mole of NaCl weighs 58.44 g.

A 2.5 M solution is 2.5 moles per liter (Molarity is just the number of moles per liter).

Therefore, 0.5 L would contain 1.25 mol. Hence, you would need 1.25 ×58.44 g = 73 g.

The molarity of a resulting solution is given by

Mole fraction of third component is 1 - (0.6 + 0.32) = 0.08.
Since ethyl chloride hashighest mole fraction so it is the solvent.