Question

A solution is made by adding 32.4 mL of concentrated nitric acid ( 70.4 wt% , density 1.42 g/mL ) to some water in a volumetric flask, and then adding water to the mark to make exactly 250 mL of solution. Calculate the concentration of this solution in molarity.

Answer 1

Answer: 2.06 M

Step-by-step explanation: Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Given : 70.4 g of
`HNO_3` is dissolved in 100 g of solution.

Density of solution = 1.42 g/ml

Volume of solution =
`(mass)/(density)=(100g)/(1.42g/ml)=70.42ml`

`Molarity=(n* 1000)/(V_s)`

where,

n= moles of solute

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}=(70.4g)/(63g/mol)=1.12moles`

`V_s` = volume of solution = 70.42 ml

`Molarity=(1.12* 1000)/(70.42)=15.9M`

According to the neutralization law,

`M_1V_1=M_2V_2`

where,

`M_1` = molarity of stock
`HNO_3` solution = 15.9 M

`V_1` = volume of stock
`HNO_3` solution = 32.4 ml

`M_2` = molarity of final
`HNO_3` solution = ?

`V_2` = volume of final
`HNO_3` solution = 250 ml

`15.9* 32.4=M* 250`

`M=2.06M`

Therefore, the concentration final solution is 2.06 M.

Answer 2

Molarity of the solution = 2.056 M

Step-by-step explanation:

Mass of nitric acid = Volume × Density

= 32.4 mL × 1.42 g/mL= 46.008 g

Nitric acid is 74% by weight.

Mass of
`HNO_3 = (70.4)/(100) * 46.008 = 32.39 g`

`Moles\;of\;HNO_3=(Mass\;of\;HNO_3)/(Molar\;Mass\;of\;HNO_3)`

Molar mass of
`HNO_3` = 63.01 g/mol

`Moles\;of\;HNO_3=(32.39)/(63.01)=0.5140\;mol`

Volume of the solution = 250 mL = 0.250 L

`Molarity=(Moles)/(Volume)`

=
`(0.5140)/(0.250) =2.056\;M`