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Murljashka [212]
3 years ago
6

Find the molarity of 186.55 g of sugar (C12H22O11) in 250. mL of water.

Chemistry
1 answer:
Anna [14]3 years ago
6 0

Answer:

The molarity of this sugar solution in water is 2.18 M

Explanation:

Step 1: Data given

Mass of sugar (C12H22O11) = 186.55 grams

Molar mass of C12H22O11 = 342.3 g/mol

Volume of water = 250.0 mL = 0.250 L

Step 2: Calculate moles sugar

Moles sugar = mass sugar / molar mass sugar

Moles sugar = 186.55 grams / 342.3 g/mol

Moles sugar = 0.545 moles

Step 3: Calculate molarity of the sugar solution

Molarity = moles sugar / volume of water

Molarity = 0.545 moles / 0.250 L

Molarity = 2.18 MThe molarity of this sugar solution in water is 2.18 M

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86 elections so 43 orbitals
6 0
3 years ago
What is the molarity when 0.181 moles of MgNO3)2 are dissolved in enough water to make 0.750 L solution?
eimsori [14]

Answer:

Molarity = 0.24 M

Explanation:

Given data:

Number of moles of Mg(NO₃)₂ = 0.181 mol

Volume of solution = 0.750 L

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

by putting values,

Molarity = 0.181 mol / 0.750 L

Molarity = 0.24 M

4 0
2 years ago
(02.01 MC)
Rina8888 [55]

Answer:

C. Particle size

Explanation:

The sand, which has smaller particles, will go through the sieve, while the rice (with a larger particle size) will not

4 0
2 years ago
If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be
lara31 [8.8K]

Answer:

Ca₃(AsO₃)₂

Explanation:

Sodium arsenite, with the chemical formula Na₃AsO₃, is formed  by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.

Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.

4 0
3 years ago
30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at
Zanzabum

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

6 0
2 years ago
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