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Virty [35]
3 years ago
12

An aqueous solution of iron(II) sulfate (FeSO4) is prepared by dissolving 2.00 g in sufficient deionized water to form a 200.00

mL solution. Calculate the molarity of the solution.
Chemistry
1 answer:
ElenaW [278]3 years ago
3 0

Answer:

0.066mol/dm^3

Explanation:

The molarity is simply calculating the number of moles in 1L or 1000ml

To get this, we need to know the amount in grammes that would be present in 1L.

Since 2g is sufficient for 200ml, then for 1000ml, the amount sufficient would be 5 * 2 = 10g

Now to get the number of moles needed, we need to know the molar mass of the compound. That is molar mass of FeSO4 = 56 + 32 + 4(16) = 152g/mol

The number of moles is thus 10/152 = 0.066 mol/dm^3

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How many atoms of hydrogen are found in the compound, NH4C2H302?
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7 hydrogen atoms

<h2>Explanation:</h2>

N<em><u>H4</u></em>C2<em><u>H3</u></em>02

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Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you
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<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (methanol) = 0.135 g

M_{solute} = Molar mass of solute (methanol) = 32 g/mol

W_{solvent} = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m

Hence, the molality of the solution is 0.11 m

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