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3 years ago

A solution of a sugar with the chemical formula (c12h22o11) is prepared by dissolving 8.45 g in 250.0 ml of water at 25

2 answers:

6 0

A. Concentration in terms of molarity

First, calculate for the molar mass of the given solution, C12H22O11.
m = 12(12) + 22(1) + 11(16) = 232
Calculate for the number of moles.

n = 8.45 g / 232 g/mol = 0.036422 moles

Divide the determined number of moles by volume of solution in L.
V(solution) = (250 mL)(1 L/1000 mL) = 0.25 L

The concentration in molarity is determined by dividing the number of moles by the volume of solution in liters.
M = 0.03622 moles / 0.25 L = 0.146 moles/L

b. Concentration in terms of molality
Calculate the mass of the solvent in kilogram given the volume and its density.
m(solvent) = (1.21 g/cm3)(250 mL)(1 kg/1000 g) = 0.3025 kg
Divide the calculated number of moles by the mass of the solvent in kilograms.

molality = 0.036422 moles/0.3025 kg = 0.12 molal

c. Concentration in terms of weight percent.
The total weight of the solution is equal to 302.5 g.

weight percent of solute = (8.45 g / 302.5 g)(100%) = 2.79%

d. Weight percent in ppm.
weight percent of solute x (10000 ppm/1%)
= (2.79%)(10000 ppm/1%) = 27933.88 ppm

tigry1 [53]3 years ago

6 0

Answer : The molarity, molality, weight percent and ppm are, 0.0987 mole/L, 0.0839 mole/Kg, 2.79% and 27933.8843 respectively.

Solution : Given,

Mass of solute $(C_{12}H_{22}O_{11})$ = 8.45 g

Molar mass of solute $(C_{12}H_{22}O_{11})$ = 342.3 g/mole

Volume of solvent (water) = volume of solution = 250 ml

Density of solution = $1.21g/cm^3=1.21g/ml$

Calculation for molarity

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$Molarity=\frac{8.45g\times 1000}{342.3g/mole\times 250ml}=0.0987mole/L$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=Density\times Volume=1.21g/ml\times 250ml=302.5g$

Now we have to calculate the mass of solvent.

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}=302.5-8.45=294.05g$

Calculation for molality

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in g}}$

$Molality=\frac{8.45g}\times 1000}{342.3g/mole\times 249.05g}=0.0839mole/Kg$

Calculation for weight percent

$weight\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{8.45g}{302.5g}\times 100=2.79\%$

Calculation for ppm

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6=\frac{8.45g}{302.5g}\times 10^6=27933.8843$

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How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt

Citrus2011 [14]

There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

$(NH_{4})2CO_{3}$

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

$N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g$

We have that the molecular mass of the substance will be:

$MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}$

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

$mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}$

$mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol$

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

$M=\frac{moles_{solute}}{volume_{solution}}$

$M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}$

Now, substituting and calculating, we have:

$volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL$

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

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3 years ago

Distinguish between sugar and non-sugar with examples.

kap26 [50]

Answer:

Reducing sugars are sugars where the anomeric carbon has an OH group attached that can reduce other compounds. Non-reducing sugars do not have an OH group attached to the anomeric carbon so they cannot reduce other compounds. ... Maltose and lactose are reducing sugars, while sucrose is a non-reducing sugar

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3 years ago

The electrical conductivities of the following 0.100 M solutions were measured in an apparatus that contained a light bulb as th

Rufina [12.5K]

Answer:

Al(NO₃)₃ > KI > HF > CH₃OH

Explanation:

The electrical conductivities of the solutions will depend on the concentration of ions in solution.

Al(NO₃)₃ solution contains 0.1 M of Al³⁺ ions and 0.3 M of NO₃⁻ ions

KI solution contains 0.1 M of K⁺ ions and 0.1 M of I⁻ ions

HF solution contains less than 0.1 M of H⁺ ions and less then 0.1 M of F⁻ ions, because the HF acid will not dissociate completely

CH₃OH practically it does not dissociate, so in the solution will not be electrical conductive (comparative with the other solutions)

The solutions in order of decreasing intensity of the bulb are ranked as following:

Al(NO₃)₃ > KI > HF > CH₃OH

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3 years ago

Magnalium alloys contain 70% Al and 30.0% Mg by mass. How many grams of H2(g) are produced in the reaction of a 0.710g sample of

Elis [28]

Answer:

H2 produced = 0.4235g

Explanation:

Equations for reaction with Hcl

Aluminium

2Al + 6 Hcl ------------ 2Alcl3 + 3H2

Magnesium

Mg + 2Hcl -------- Mgcl2 + H2

Aluminium = 70% of 0.710g of sample

sample contains (70 x 0.710)/100 = 0.497g of aluminium

Magnesium = 30% of 0.710g of sample

sample contains(30 x 0.710)/100 = 0 .213g of magnesium

Moles = mass/ molecular mass

Moles of aluminium in sample = 0.497/27 = 0.0184

2 moles of aluminium gives 3 moles of H2

No of moles of H2 from reaction with aluminium = (2 x0.0184)/3

= 0.0123 moles

1 mole of H2 = 2g therefore mass of H2 produced = 0.0123 x 2 = 0.0246g

Moles of magnesium in sample = 0.213/24 = 0.008875

1 mole of mg gives 1 mole of H2

No of moles of H2 from reaction with magnesium = 0.008875 x 1

= 0.008875

1 mole of H2 = 2g therefore mass of H2 produced = 0.008875 x 2

= 0.01775g

Ttal mass of H2 = 0.0246 +0.01775 = 0,04235g

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3 years ago

The weight of Patterson on the moon will be ..... This is 6th grade science by the way

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Answer:The weight of Patterson on the moon would be less.

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3 years ago