Question

what quantity of NaCl is required for preparing 200ml of 0.9%solution​

Answer 1

For example, 0.9% NaCl contains 0.9 g NaCl per 100 ml of solution or 9 g NaCl/l. This can be converted to molarity by dividing by molecular weight: 0.9% NaCl = (9g/l)/(58.5g/mole) = 0.15 M NaCl. Mole: A mole (or mol) of a substance is simply Avogadro's number particles of that substance.

Answer 2

Given:

A certain amount of NaCl is present.

To Find:

An amount of NaCl is required for preparing 200ml of 0.9%solution.

Solution:

To find the amount of NaCl is required for preparing 200ml of 0.9%solution we will follow the following steps:

As we know,

0.9% Means 0.9 gram of NaCl is present in 100ml of solution.

So, the amount of NaCl requires to prepare 1 ml of solution

$= \frac{0.9}{100}$

Now,

The amount of NaCl requires to prepare 200 ml of solution =

$\frac{0.9}{100} \times 200 = 1.8 \: gram$

Henceforth, the amount of NaCl required for preparing 200ml of 0.9%solution is 1.8 grams.