Question

50 g of 15% (w/w)sodium hydroxide solution is prepared in a laboratory .calculate the volume of water utilized if the density of water is 0.9g/cc at the experimental condition

Answer 1

Answer:- 47.2 mL

Solution:- 15%(w/w) sodium hydroxide solution means 15 grams of sodium hydroxide present in 100 grams of solution. Let's calculate the grams of sodium hydroxide present in 50 grams of the solution as:

$50gsolution(\frac{15gNaOH}{100gsolution)})$

= 7.5 g NaOH

since, mass of solution = mass of solute + mass of solvent

So, mass of solvent = mass of solution - mass of solute

mass of solvent = 50 g - 7.5 g = 42.5 g

The solvent is water and it's density is given as $\frac{0.9g}{cc}$ where cc stands for cubic centimeter and we know that, 1 cc = 1 mL

So, the density of water is $\frac{0.9g}{mL}$ .

$volume=\frac{mass}{density}$

$42.5g(\frac{1mL}{0.9g})$

= 47.2 mL

So, to make 50 g of 15%(w/w) sodium hydroxide solution, 47.2 mL of water are utilized.

Answer 2

Given:

50 g of 15% (w/w)sodium hydroxide solution

Density of water = 0.9g/cc

Solution:

From given,

15 grams of sodium hydroxide is present in 100 grams of solution.

To calculate the grams of sodium hydroxide present in 50 grams of the solution,

50g solution ( 50g NaOH/100 solution)

We get,

7.5 g NaOH

By property,

Mass of solution = mass of solute + mass of solvent

And so,

Mass of solvent = mass of solution - mass of solute

Therefore,

Mass of solvent = 50 g - 7.5 g

Mass of solvent = 42.5 g

As solvent's density is expressed in cc,

It's density is given by 0.9g / cc

Where,

cc - cubic centimeter

1 cc = 1 mL

42.5 ( 1 / 0.9) = 47.2 mL

Hence,

The density of water is = 47.2 mL

So, to make 50 g of 15%(w/w) sodium hydroxide solution, 47.2 mL of water is needed under experimental conditions.