Chemistry, asked by Satvik1476, 9 months ago


How much water will be required to prepare 20% aqueous salt solution by adding 10 g of common
salt?​

Answers

Answered by yashikapandya
6

Answer:

How much water will be required to prepare 20% aqueous salt solution by adding 10 g of common salt.

Answered by HrishikeshSangha
0

40 g of water will be required to prepare 20% aqueous salt solution by adding 10 g of common salt.

Given,

Weight by weight percentage=20%

Mass of common salt(NaCl)=10 g.

To find,

the amount of water required to prepare 20% aqueous salt solution by adding 10 g of common salt.

Solution:

  • The concentration term here referred to is weight by weight percentage.
  • Weight by weight percentage of a substance in a mixture or a solution is defined as the mass of that substance in 100 g of solution or mixture.
  • Weight by weight percentage of a substance is temperature independent.
  • Weight by weight percentage is also referred as mass by mass percentage.
  • It is calculated as:                          
  • %w/w=(Mass of the substance/Mass of the mixture or solution) x 100.

Let the mass of solvent required to prepare the solution be m.

Mass of the solution=Mass of solvent+Mass of solute

Mass of the solution=(m+10) g.

%w/w=(Mass of the substance/Mass of the mixture or solution) x 100

20=\frac{10}{(10+m)} X100\\\frac{20}{100}=\frac{10}{(10+m)} \\\frac{1}{5}=\frac{10}{(10+m)}\\(10+m)=5X10\\10+m=50\\m=40g.

The amount of water required is 40 g.

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