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Black_prince [1.1K]
3 years ago
12

If 36.0 g of NaOH (MM = 40.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the

concentration of NaOH in the resulting solution
Chemistry
1 answer:
Kobotan [32]3 years ago
8 0

Answer:

Molarity of NaOH = 1.8 M.

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Volume = 500 mL

Molarity of NaOH =?

Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 36 / 40

Mole of NaOH = 0.9 mole

Next, we shall convert 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.9 mole

Volume = 0.5 L

Molarity of NaOH =?

Molarity = mole / Volume

Molarity of NaOH = 0.9 / 0.5

Molarity of NaOH = 1.8 M

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Read 2 more answers
Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl
tensa zangetsu [6.8K]

Answer:

[SO_2Cl_2] = 0.09983 M

Explanation:

Write the balance chemical equation ,

SO_2Cl_2((g) = SO_2(g) + Cl_2(g)

initial concenration of SO_2Cl_2((g)  =0.1M

lets assume that degree of dissociation=\alpha

concenration of each component at equilibrium:

[SO_2Cl_2] = 0.1-0.1\alpha

[SO_2] = 0.1\alpha

[Cl_2] = 0.1\alpha

Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}

Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}

as \alpha is very small then we can neglect  1-\alpha

therefore ,

Kc ={0.1\alpha \times \alpha}

\alpha =\sqrt{\frac{Kc}{0.1}}

\alpha = 1.73 \times 10^{-3}

Eqilibrium concenration of [SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173

[SO_2Cl_2] = 0.09983 M

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