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Oksana_A [137]
3 years ago
13

A solution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.15 ∘c. what is the molal concentration of glucose in this solution

? assume that the freezing point of pure water is 0.00 ∘c.
Chemistry
1 answer:
Reil [10]3 years ago
7 0
<span>1.16 moles/liter The equation for freezing point depression in an ideal solution is ΔTF = KF * b * i where ΔTF = depression in freezing point, defined as TF (pure) â’ TF (solution). So in this case ΔTF = 2.15 KF = cryoscopic constant of the solvent (given as 1.86 âc/m) b = molality of solute i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1. Solving for b, we get ΔTF = KF * b * i ΔTF/KF = b * i ΔTF/(KF*i) = b And substuting known values. ΔTF/(KF*i) = b 2.15âc/(1.86âc/m * 1) = b 2.15/(1.86 1/m) = b 1.155913978 m = b So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
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Answer:

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Explanation:

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3 years ago
A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what
Charra [1.4K]
1) mass composition

N: 30.45%
O: 69.55%
   -----------
   100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

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3 0
2 years ago
A buffer solution contains 0.120M acetic acid and 0.150M sodium acetate.
Dvinal [7]
A. M x L = moles. 
<span>b. CH3COOH + NaOH ==> CH3COONa + H2O </span>
<span>I...6 mmols....0.......7.5 mmoles </span>
<span>C... 0........0.51 mmols..0 </span>
<span>E...6-0.511 ....0.......7.5+0.511 </span>

<span>I stands for initial </span>
<span>C stands for change. </span>
<span>E stands for equilibrium. </span>
<span>Just divide mmoles by 1000 to convert to moles. I work in mmoles because I get tired of writing those zeros. </span>

<span>c. done as in b.</span>
8 0
3 years ago
Plz help me out!!!!!
Lelu [443]

Answer:

i am not 100% sure but im pretty sure there is.

Explanation:

7 0
2 years ago
If 28 grams of N reacts completely with 12 grams of H2, then how many
Bogdan [553]

Answer:

Mass of NH₃ produced = 34 g

Explanation:

Given data:

Mass of nitrogen = 28 g

Mass of Hydrogen = 12 g

Mass of NH₃ produced = ?

Solution:

Chemical equation:

N₂ +  3H₂    →   2NH₃

Moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 28 g/ 28 g/mol

Number of moles = 1 mol

Moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 12 g/ 2 g/mol

Number of moles = 6 mol

Now we will compare the moles of hydrogen and nitrogen with ammonia.

                            H₂              :               NH₃

                            3                :                2

                            6                :             2/3×6 = 4 mol

                           N₂              :                NH₃

                            1                :                 2

Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.

Mass of ammonia produced:

Mass = number of moles × molar mass

Mass =  2 mol  ×  17 g/mol

Mass = 34 g

                     

5 0
3 years ago
Read 2 more answers
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