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Zolol [24]
4 years ago
12

A solution is prepared by dissolving 10.0 g NaCl in 600 g of water. What is its molality? (Note: the molar masses are NaCl = 58.

5 g/mol; H2O = 18.0 g/mol)
A.
0.892m NaCl
B.
0.285m NaCl
C.
5.13m NaCl
D.
1.81m NaCl
Chemistry
1 answer:
xxMikexx [17]4 years ago
3 0

Answer:

Option B. 0.285m NaCl

Explanation:

To find the molality, first let find the number of mole of NaCl in the solution. This is illustrated below

Molar Mass of NaCl = 58.5 g/mol

Mass of NaCl = 10g

Number of mole of NaCl = Mass of NaCl / molar Mass of NaCl

Number of mole of NaCl = 10/58.5 = 0.1709mol

Mass of water = 600g = 600/1000 = 0.6kg

Molality = mole of solute / mass of solvent

Molality = 0.1709mol/0.6kg

Molality = 0.285mol/Kg

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3 years ago
Solid potassium chlorate decomposes into solid potassium chloride and oxygen gas. If 2.00 g potassium chlorate decomposes in a c
abruzzese [7]

Answer : The final chamber pressure is 0.746 atm.

Explanation:

First we have to calculate the moles of KClO_3.

Molar mass of KClO_3 = 122.5 g/mole

\text{ Moles of }KClO_3=\frac{\text{ Mass of }KClO_3}{\text{ Molar mass of }KClO_3}=\frac{2.00g}{122.5g/mole}=0.0163moles

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that,

As, 2 moles of KClO_3 react to give 3 moles of O_2

So, 0.0163 moles of KClO_3 react to give \frac{3}{2}\times 0.0163=0.0244 moles of O_2

Now we have to calculate the pressure of gas.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = ?

V = volume of gas = 0.800 L

T = temperature of gas = 25.0^oC=273+25.0=298K

R = gas constant = 0.0821 L.atm/mole.K

n = number of moles of gas = 0.0244 mole

Now put all the given values in the ideal gas equation, we get:

P\times (0.800L)=0.0244mole\times (0.0821L.atm/mole.K)\times (298K)

P=0.746atm

Therefore, the final chamber pressure is 0.746 atm.

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4 years ago
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Explanation:

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Explanation:

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