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Reika [66]
3 years ago
15

Calculate the molality of 2.0 M MgCl2 solution. The density of the solution is 1.127 g/mL. (The molar mass of MgCl2 = 95.211 g/m

ol and the molar mass of H2O is 18.02 g/mol.)
Chemistry
2 answers:
Elan Coil [88]3 years ago
8 0

The answer is 2.135 mol/Kg

Given that molarity is 2M, that is, 2 moles in 1 liter of solution.

Density of solution is 1.127 g/ml

Volume of solution is 1L or 1000 ml

mass of solution (m) = density × volume

m₁ = density × volume = 1.127 × 1000 = 1127 g

mass of solute, m₂ = number of moles × molar mass

m₂ = 2 × 95.211

m₂ = 190.422 g

mass of solvent = m₁ - m₂

= 1127 - 190.422

= 936.578 g

= 0.9366 Kg

molality = number of moles of solute / mass of solvent (in kg)

= 2 / 0.9366

= 2.135 mol/Kg

denis23 [38]3 years ago
4 0
The answer is 2.135 mol/Kg
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If the hydroxide ion concentration of a solution is 3.26 x 10-6 M, is it an acidic or a basic solution?
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Explanation:

We were given that hydroxide ion concentration [OH-] as 3.26 x 10-6 M

But We know thar

[OH-] [H+] = 1 × 10^-14

To get the Hydrogen ion [H+] concentration, we have that

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But, pH = - log [H+]

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pH=8.51

when  pH > 7, The solution is basic, therefore a solution with pH =8.51 is basic.

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Starting with the stock solution of 6.0 M, how many milliliters of 6.0 M sulfuric acid are needed to make 450 mL of 1.2 M soluti
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<u>Answer:</u> The volume of stock solution needed is 90 mL

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

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M_1\text{ and }V_1 are the molarity and volume of the stock sulfuric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted sulfuric acid solution

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