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den301095 [7]
3 years ago
10

What mass of sucrose c12h22o11 is needed to make 300 ml of a 0.50m solution?

Chemistry
1 answer:
zepelin [54]3 years ago
5 0
Use the concentration to obtain the moles. I am assuming you mean to write capital M. because little m means molality. 

So, first convert the ml into Liters and then into moles, then moles to grams using the molar mass (just adding the values of each atom from the periodic table. )

Molar mass= 12 (12.0) + 22 (1.01)+ 11 (16.0)= 342 grams/mole

300 ml (1 liter/ 1000 mL) x (0.50 moles/ 1 Liter) x (342 grams/ 1 mole)= 51.3 grams


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Option D. pH= 1.3 strong acid

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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
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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.

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