Question

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

Answer 1

Answer: 41.5 mL

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

`Molarity=(n)/(V_s)`

where,

n = moles of solute

`V_s` = volume of solution in L

Given : 59.4 g of
`H_2SO_4` in 100 g of solution

moles of
`H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=(59.4g)/(98g/mol)=0.61`

Volume of solution =
`\frac{\text {mass of solution}}{\text {density of solution}}=(100g)/(1.83g/ml)=54.6ml`

Now put all the given values in the formula of molality, we get

`Molality=(0.61* 1000)/(54.6ml)=11.2M`

To calculate the volume of acid, we use the equation given by neutralisation reaction:

`M_1V_1=M_2V_2`

where,

`M_1\text{ and }V_1` are the molarity and volume of stock acid which is
`H_2SO_4`

`M_2\text{ and }V_2` are the molarity and volume of dilute acid which is
`H_2SO_4`

We are given:

`M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL`

Putting values in above equation, we get:

`11.2* V_1=0.30* 1550\\\\V_1=41.5mL`

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid