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Stels [109]
3 years ago
9

How would you prepare a 1 L solution of 3 M MgO?

Chemistry
1 answer:
OleMash [197]3 years ago
8 0

Answer:

You need to measure 120.9132 grams of MgO and add to 1 liter of distilled water in a beaker.

Explanation:

Capital M stands for molarity, or moles/liter. So 3 M MgO means 3 moles per liter of MgO.

1 L × \frac{3 moles MgO}{1 L MgO} = 3 moles MgO

This means you need to measure 3 moles MgO to put into 1 Liter of water.

But you don't know how much that is, so you need to convert moles to grams.

First you need to find how many grams are in 1 mole of MgO.

You can search the molar mass of MgO online to get: 40.3044 g

But if you don't have the internet, use the periodic table to add the masses of Mg and O: 24.305 g + 15.999 g = 40.3044 g

Then write the fraction so that the units cancel out, so you get grams by itself on the top of the fraction.

1 L × \frac{3 moles MgO}{1 L MgO} × \frac{40.3044 g MgO}{1 mole MgO} = 120.9132 g MgO

You need to measure 120.9132 grams of MgO and add to 1 liter of distilled water.

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Suppose a salt and a glucose solution are separated by a membrane that is permeable to water but not to the solutes. the nacl so
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1)

<span>m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = </span><span>58.5 g/mole

Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

</span>V(H2O) = 250ml = 250g = 0.25 kg<span>

</span><span>molality of NaCl:
</span><span>
n(NaCl)=m/M=1.95/58.5= 0.033 mole

</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation

Since NaCl dissociates into 2 ions in solution:
<span>                                        
</span>milliosmolality of NaOH = 132 x 2 = 264  osmol<span>es/kg
</span>
2)

m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole

Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

V(H2O) = 250ml = 250g = 0.25 kg

molality of glucose:

n(gl)=m/M=9/180= 0.05 mole

molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg

milimolality of glucose = 0.132/0,001 = 200 mmole/kg

milliosmolality of glucose = milimolality x N of ions formed in dissociation

Since glucose does not dissociate, milimolality and milliosmolality are same:
                                        
milliosmolality of glucose = 200 osmoles/kg

3)

The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.

In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.

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