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A saline solution, NaCl in water, is 0.92 % (m/v). How many grams of NaCl are required to prepare 35.0 mL of this solution?

Sagot :

oladayoademosu

Answer:

0.322 g

Explanation:

Since our concentration is given in mass per volume percent % (m/v)

% (m/v) = mass of solute in g/volume of solution in mL × 100%

Since % (m/v) = 0.92 and volume of the required solution is 35.0 mL, we find the mass of NaCl from

mass of NaCl = % (m/v)/100 % × volume of solution

= 0.92 % (m/v)/100 %× 35 mL

= 0.0092 × 35

= 0.322 g

The study of a chemical is called chemistry.

The correct answer is 0.322 g

The saline is said that the amount of salt present in the solution. The formula used to solve the question is as follows:-

[tex]\frac{m}{v} = \frac{mass \ of\ solute}{volume\ of\ solution} * 100[/tex]

The data is given as follows:-

  • 0.92 %(m/v)
  • Volume is 35.0 mL,

Mass of NaCl is as follows:-

 [tex]\frac{0.92}{100} * 35 mL[/tex]

[tex]0.0092 * 35\\= 0.322 g[/tex]

Hence, the correct answer is 0.322g.

For more information, refer to the link:-

https://brainly.com/question/14698383

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