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3 years ago

A substance has a solubility of 350 ppm.how many grams of the substance are present in 1.0l of a saturated solution

Chemistry

2 answers:

Kazeer [188]3 years ago

8 0

Answer:

0.350 grams of the substance are present in 1.0l of a saturated solution

Explanation:

Parts per million (ppm), is a unit of measurement for concentration. Refers to the number of units of the substance per million units of the set.

So ppm means parts per million, that is, the mg (milligrams) in a kg of solution.

Considering that the density of a substance is 1000 kg / m3 (and 1 m3 is 1000 L), 1 kg of solution has a volume of approximately 1 liter, ppm also represents the mg of a substance in a liter of solution. That is, 1 ppm = 1 mg / L.

So in 1 L there are 350 mg of substance present. Being 1 g = 1000 mg, then 350 mg is 0.350 g.

0.350 grams of the substance are present in 1.0l of a saturated solution

sergeinik [125]3 years ago

6 0

Since the given solubility is 350 ppm, convert it first with fraction of solubility. by dividing the solubility with 10^6
S = 350 / 10^6
s = 3.5 x 10^-4
the multiply it to the total solution to calculate the amount of substance present
m = ( 3.5 x 10^-4 ) ( 1.01 )
m = 3.535 x 10^-4 g of the substance present

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A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many gra

denis-greek [22]

Answer:

0.52 g of chromium(II) hydroxide, Cr(OH)2.

Explanation:

We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.

This can be obtained as follow:

Molarity of CrCl2 = 0.258 M

Volume = 35.9 mL = 35.9/1000 = 0.0359 L

Mole of CrCl2 =?

Molarity = mole /Volume

0.258 = mole of CrCl2 /0.0359

Cross multiply

Mole of CrCl2 = 0.258 x 0.0359

Mole of CrCl2 = 0.0093 mole

Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.

This can be obtained as follow:

Molarity of KOH = 0.338 M

Volume = 35.8 mL = 35.8/1000 = 0.0358 L

Mole of KOH =.?

Molarity = mole /Volume

0.338 = mole of KOH /0.0358

Cross multiply

Mole of KOH = 0.338 x 0.0358

Mole of KOH = 0.0121 mole.

Next, we shall write the balanced equation for the reaction. This is given below:

2KOH + CrCl2 → Cr(OH)2 + 2KCl

From the balanced equation above,

2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

2 mole of KOH reacted with 1 mole of CrCl2.

Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.

From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.

Therefore, KOH is the limiting reactant.

Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.

In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.

The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:

From the balanced equation above,

2 mole of KOH reacted to produce 1 mole of Cr(OH)2.

Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.

Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.

This is illustrated below:

Mole of Cr(OH)2 = 0.00605 mole

Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol

Mass of Cr(OH)2 =..?

Mole = mass /Molar mass

0.00605 = mass of Cr(OH)2/86

Cross multiply

Mass of Cr(OH)2 = 0.00605 x 86

Mass of Cr(OH)2 = 0.52 g

Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.

6 0

3 years ago