Chemistry: How many moles of nacl is needed to prepare 67.3g of 21% solution

dynamic equilibrium is the answer....

How many moles of nacl is needed to prepare 67.3g

sarbjit879

11.01.2022

The total number of moles for making 21% NaCl solution is : 0.1997mol .

Explanation:

Given :

21% solutionWeight = 67.3g

Weight of

If a solution is ''% , it means : 'x' g of the solute is present in 100ml of solution.

Here the solvent is water.

∴

21% Solution :

$\frac{21gNaCl}{100mlH_{2}O}$

Since the volume of water doesn't change much as the salt dissolves or ionizes ,

we can assume that we have used 100g of $H_{2}O$ has been used for making this solution.

thus , EVERY 121g

of the solution , contains 21g

thus 67.3g will have :

$\frac{67.3}{121} *21//=11.68g$

Thus , number of moles :

$\frac{total.weight}{weight.per.mole}//=\frac{11.68}{58.5}//=0.1997mol$

Answer : 0.1997mol

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