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DaniilM [7]
3 years ago
8

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with di

stilled water. What is the final molarity of the dilute solution?
Chemistry
1 answer:
Zina [86]3 years ago
6 0

Answer:

0.102 M.

Explanation:

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

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M1) = 5.103 M

Volume of diluted solution (V2) = 250 mL

Molarity of diluted solution (V2) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M1V1 = M2V2

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

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Answer:

m_{H2SO4 = 61.25 g

m_{Na2SO4} = 88.75 g

Explanation:

m_{NaOH} = \frac{200 . 25 }{100} = 50 g

⇒ n_{NaOH} = \frac{50}{40} = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

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⇒  n_{H2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ m_{H2SO4} = 0.625 × 98 = 61.25 g

    n_{Na2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒m_{Na2SO4} = 0.625 × 142 = 88.75 g

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3 years ago
Question
andreev551 [17]

Answer:

Sln

n=m/mr

n=25/100

n=0.25mole of Caco3

Malality =number of moles/volume (divided by number of moles both sides)

volume =Malality /number of moles

v=0.125/0.25

v=0.500L

I hope this help

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