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Sonbull [250]
3 years ago
5

what is molarity of a sodium hydroxide solution made by combining 2.0 L of 0.60 M NaOH with 495 mL of 3.0 M NaOH? Assume the vol

umes of the two solutions to be additive
Chemistry
1 answer:
hammer [34]3 years ago
8 0

Answer:

Molarity of the sodium hydroxide solution is 1.443 M/L

Explanation:

Given;

0.60 M concentration of NaOH contains 2.0 L

3.0 M concentration of NaOH contains 495 mL

Molarity is given as concentration of the solute per liters of the solvent.

If the volumes of the two solutions are additive, then;

the total volume of NaOH = 2 L + 0.495 L = 2.495 L

the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M

Molarity of NaOH solution = 3.6 / 2.495

Molarity of NaOH solution = 1.443 M/L

Therefore, molarity of the sodium hydroxide solution is 1.443 M/L

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Answer:

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Explanation:

The given values are:

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⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

⇒  Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}

⇒                                             =1.5\times 10^{-3}  

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⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

⇒                         =0.60 \ g

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