In: Chemistry

A solution is 3.24 M NH_{3} (MM 17.04 g/mol) in water
(MM 18.02 g/mol). The density of the resulting solution is 0.974
g/mL. What is the mole fraction, mass percent and molality of
NH_{3} in the solution?

The solution is 3.24 M in NH_{3}. Consider 1.00 L of the
solution.

Moles of NH_{3} in 1.00 L solution = (1.00 L)*(3.24 M) =
(1.00 L)*(3.24 mol/L) = 3.24 moles.

Molar mass of NH_{3} = 17.04 g/mol; therefore, mass of
NH_{3} corresponding to 3.24 moles = (3.24 mole)*(17.01
g/mol) = 55.2096 g.

Density of the solution is 0.974 g/mL; therefore, mass of 1.00 L solution = (1.00 L)*(1000 mL/1 L)*(0.974 g/mL) = 974.0000 g.

Mass of water in the solution = (mass of solution) – (mass of
NH_{3}) = (974.0000 g) – (55.2096 g) = 918.7904 g =
(918.7904 g)*(1 kg/1000 g) = 0.9187904 kg

Moles of water corresponding toe 918.8876 g water = (mass of water)/(molar mass of water) = (918.7904 g)/(18.02 g/mol) = 50.9872 moles.

Mole fraction of NH_{3} in the solution = (moles of
NH_{3})/(total number of moles) = (3.24 moles)/(3.24 moles
+ 50.9872 moles) = 3.24/54.2272 = 0.0597 ≈ 0.06 (ans).

Mass percent NH_{3} = (mass of NH_{3})/(mass of
solution)*100 = (55.2096 g)/(974.0000 g)*100 = 5.6683% ≈ 5.67%
(ans).

Molality of NH_{3} = (moles of NH_{3})/(kg of
water) = (3.24 moles)/(0.9187904 kg) = 3.5263m ≈ 3.53m (ans).

A solution is 5.66 % by mass NH3 (MM 17.04 g/mol) in water (MM
18.02 g/mol). The density of the resulting solution is 0.974 g/mL.
What is the mole fraction, molarity and molality of NH3 in the
solution?

A solution has a 0.102 mole fraction of CaCl2 (MM
110.98 g/mol) in water (MM 18.02 g/mol). The density of the
solution is 1.05 g/mL. What is the molality, mass percent and
molarity of CaCl2 in the solution?

A 0.148-M solution of X (Mm = 177 g/mol) has a density of 1.156
g/ml. What is the molality of X in the solution?

15.00 g of aluminum sulfide (150.1 g/mol) and 10.00 g of water
(18.02 g/mol) react until the limiting reactant is used up.
Calculate the mass of H2S (34.08 g/mol) that can be produced from
these reactants. Notice that you will need to balance the reaction
equation.
___Al2S3(s)+ ___H2O > ___Al(OH)3(s)+ ___H2S(g)
13.89 g d. 9.456 g
10.21 g e. 1.108 g
19.67 g

1.8 grams of ammonia, NH3 (M=18 g/mol), is dissolved in 10L of
water. Which best approximates the pH of the solution? Please
explain/show work.
The answer is 7-11.9, but I got 12 and was wondering why it's
less than that.

If the density of water is 1g/mL and the density of a 0.16 M HCl
solution is 1.023 g/mL, how do I use this information to determine
the mass of both water and HCl in one liter of the solution?
***Cancel this question! I figured it out!

When 150.0 mL of a 0.200 M Ca(NO3)2 (MM = 164.088 g/mol)
solution is added to 200.0 mL of a 0.150 M K3PO4 (MM = 212.27
g/mol) solution, what mass of calcium phosphate (MM = 310.17 g/mol)
will precipitate?
a. 9.31 g
b. 3.10 g
c. 2.91 g
d. 1.40 g
e. 4.65 g

You are provided with a 0.1 mM solution of proflavine (MW:
209.25 g/mol), a 6 mM solution of DNA, and solid NaCl (MW: 58.44
g/mol). How many μL of the proflavine solution would you need to
make a 25 mL of the following stock solution: 1 µM proflavine, 100
mM NaCl, and 60 µM DNA?

A buffer is prepared by adding 3.55g NH3 (MM=17.03 g/mol) to
750.0 mL of 0.175M of HCl. What is the pH of the buffer? (Kb =
1.8*10^-5) (NO VOLUME CHANGE)

The density of a 40 wt% solution of ethanol in water is 0.937
g/mL. The density of n-butanol is 0.810 g/mL. What is the
concentration of n-butanol in ppm if you dissolve 20 μL of this
alcohol in a 40 wt% solution of Ethanol in water. The final volume
of the solution is 25 mL. Assume that the density of the ethanol
solution does not change upon dissolution of the butanol. Use the
correct number of significant figures! Hint: to...

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