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2 years ago

Calculate the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 ml solution.

1 answer:

8 0

Taking into account the definition of molarity, the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 mL solution is 2 $\frac{moles}{liter}$ .

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

<h3>Molarity in this case</h3>

In this case, you have:

number of moles= 0.2 moles
volume= 100 mL= 0.1 L

Replacing in the definition of molarity:

$molarity=\frac{0.2 mole}{0.1 L}$

Solving:

<u><em>molarity= 2 </em></u> $\frac{moles}{liter}$

Finally, the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 mL solution is 2 $\frac{moles}{liter}$ .

Learn more about molarity:

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What is the charge on an ion that has an atomic number of 16 and contains 14e-?

lora16 [44]

The answer would be 2+ since the atomic number represents how many protons are in the element. In this case, there are 16 protons, but only 14 electrons, which means there are an additional 2 protons, hence the 2+ charge on the ion.

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3 years ago

Convert each of the measurements given to the new unit stated using the factor label method (dimensional analysis). Show all wor

KengaRu [80]

For cm to m, you need to divide by 100 to get m:
26cm/100= 0.26m

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3.25km x 1000= 3250m

For ft to in, you need to multiply by 12 to get in:
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So we have:
0.26m
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6 0

3 years ago

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statuscvo [17]

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3 0

3 years ago

What mass of oxygen reacts with 4.62 g of magnesium to form magnesium oxide according to the following... 2 Mg O2 --> MgO

Goryan [66]

Answer:

3.08 g of Oxygen .

Explanation:

2Mg + O₂ = 2MgO

2 mole 1 mole

molecular weight of Mg = 24

2 mol of Mg = 48 g

48 g of Mg reacts with 32 g of oxygen

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7 0

4 years ago

"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite react

frosja888 [35]

<u>Answer:</u> The given reaction is non-spontaneous in nature.

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$3Mg(s)+Cr_2O_3(s)\rightarrow 3MgO(s)+2Cr(s)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(3\times \Delta S^o_{(MgO(s))})+(2\times \Delta S^o_{(Cr(s))})]-[(3\times \Delta S^o_{(Mg(s))})+(1\times \Delta S^o_{(Cr_2O_3(s))})]$

We are given:

$\Delta S^o_{(Mg(s))}=32.68J/K.mol\\\Delta S^o_{(Cr_2O_3(s))}=81.2J/K.mol\\\Delta S^o_{(MgO(s))}=26.94J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(3\times (26.94))+(2\times (23.77))]-[(3\times (32.68))+(1\times (81.2))]\\\\\Delta S^o_{rxn}=-50.88J/K=-0.0509kJ/K.mol$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

To calculate the standard Gibbs free energy of the reaction, we use the equation:

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy change of the reaction = 665.1 kJ/mol

T = Temperature = 298.15 K

$\Delta S^o$ = standard entropy change of the reaction = -0.0509 kJ/K.mol

Putting values in above equation, we get:

$\Delta G^o=(665.1kJ/mol)-(298.15K\times (-0.0509kJ/K.mol))=680.27kJ/mol$

As, the Gibbs free energy of the reaction is coming out to be positive, the reaction is non-spontaneous in nature.

Hence, the given reaction is non-spontaneous in nature.

4 0

4 years ago