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mario62 [17]
2 years ago
8

Sodium hydroxide is extremely soluble in water. At a certain temperature, a saturated solution contains 567 g of NaOH(s) per lit

er of solution. Calculate the molarity of this saturated NaOH(aq) solution
Chemistry
1 answer:
skad [1K]2 years ago
6 0

Answer:

14.2M

Explanation:

Amount= Concentration × Volume

But volume is 1 litre as stated in the question

And amount = mass/ molar mass

Molar mass of NaOH= 23+16+1=40gmol-1

Amount of NaOH= 567/40=14.2moles

But concentration= amount/volume

Hence molarity= 14.2/1 = 14.2M

You might be interested in
How many acidic protons are there in 0.6137 g of KHP?
andreyandreev [35.5K]

0.6137 g of KHP contains 1.086 × 10^21 acidic protons.

Number of moles of KHP = mass of KHP/molar mass of KHP

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.6137 g

Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP

Now,  1 each molecule of KHP contains 1 acidic proton.

For 0.003 moles of KHP there are; 0.003 × 1 × NA

Where NA is Avogadro's number.

So;  0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23

= 1.086 × 10^21 acidic protons.

Learn more: brainly.com/question/16672114

7 0
2 years ago
A 0.20 mol sample of MgCl2(s) and a 0.10 mol sample of KCl(s) are dissolved in water and diluted to 500 mL. What is the concentr
igor_vitrenko [27]

Answer:

1 M

Explanation:

Magnesium chloride will furnish chloride ions as:

MgCl_2\rightarrow Mg^{2+}+2Cl^-

Given :

Moles of magnesium chloride = 0.20 mol

Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:

Moles =2\times 0.20\ moles

Moles of chloride ions by magnesium chloride = 0.40 moles

Potassium chloride will furnish chloride ions as:

KCl\rightarrow K^{+}+Cl^-

Given :

Moles of potassium chloride = 0.10 moles

Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:

Moles of chloride ions by potassium chloride = 0.10 moles

Total moles = 0.40 + 0.10 moles = 0.50 moles

Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)

Concentration of chloride ions is:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{Cl^-}=\frac{0.50}{0.5}

<u> The final concentration of chloride anion = 1 M</u>

8 0
2 years ago
If one has a solution of 0.10 m silver nitrate and it is diluted by a factor of two, what is the new concentration
Angelina_Jolie [31]
Diluted by a factor of two means that we double the volume of the solution by adding an equal volume of the water.
if we diluted it by a factor of one so the new concentration = 0.1/2=0.05 M and diluted by a factor of two so, the new concentration will be 0.05/2 = 0.025 M
7 0
2 years ago
A gaseous compound is 30.4% nitrogen and 69.6% oxygen. a 6.06-gram sample of gas occupies a volume of 1.00 liter and exerts a pr
aksik [14]

A gaseous compound is 30.4% nitrogen and 69.6% oxygen by mass. A 5.25-g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 1.26 atm at -4.0°C. Which of the following is its molecular formula?  

1) NO2  

2) N3O6  

3) N2O5  

4) N2O4  

5) NO

8 0
2 years ago
Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)
steposvetlana [31]
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
      = (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
      = 176.15 kJ - 84.78 kJ = 91.38 kJ 
 





 
6 0
2 years ago
Read 2 more answers
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