Question

How many grams of hydrogen peroxide (H2O2) must be added to 1,500 mL of water to produce a concentration of 1.33 m (molal solution)? (Recall that the density of water is 1.0 gram/mL.)

0.059 g
30. g
68 g
68,000 g

Answer 1

Answer : The correct option is, 68 g

Explanation :

First we have top calculate the mass of solvent (water).

Formula used :

`Density=(Mass)/(Volume)`

Now put all the values in this formula, we get the mass of solvent (water).

`1.0g/ml=(Mass)/(1500ml)`

`\text{Mass of water}=1500g`

Now we have to calculate the mass of hydrogen peroxide.

Molality : It is defined as the number of moles of solute present in one kilogram of solvent.

In this question, the solute is hydrogen peroxide.

Formula used :

`Molality=(w_b)/(M_b* w_a)* 1000`

where,

Molality = 1.33 mole/Kg

`w_a` = mass of solvent (water) = 1500 g

`w_b` = mass of solute hydrogen peroxide = ?

`M_b` = molar mass of solute hydrogen peroxide = 34 g/mole

Now put all the given values in the above formula, we get the mass of hydrogen peroxide.

`1.33mole/Kg=(w_b)/(34g/mole* 1500g)* 1000`

`w_b=67.83g=68g`

Therefore, the grams of hydrogen peroxide will be, 68 grams.

Answer 2

You must add 68 g H_2O_2 to 1500 mL water.

Step 1. Calculate the mass of water

Mass of water = 1500 mL × 1.0 g/1 mL 1500 g = 1.50 kg

Step 2. Calculate the moles of H_2O_2.

Moles of H_2O_2 = 1.500 kg water × 1.33 mol H_2O_2/1 kg water

= 2.00 mol H_2O_2

Step 3. Calculate the mass of H_2O_2

Mass of H_2O_2 = 2.00 mol H_2O_2 × 34.01 g H_2O_2/1 mol H_2O_2

= 68 g H_2O_2