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

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 = $\frac{mass}{density}=\frac{100g}{1.42g/ml}=70.42ml$

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

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

$V_s$ = volume of solution = 70.42 ml

$Molarity=\frac{1.12\times 1000}{70.42}=15.9M$

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

$15.9\times 32.4=M\times 250$

$M=2.06M$

Therefore, the concentration final solution is 2.06 M.

Answer 2

Answer:

Molarity of the solution = 2.056 M

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 $HNO_3 = \frac{70.4}{100} \times 46.008 = 32.39 g$

$Moles\;of\;HNO_3=\frac{Mass\;of\;HNO_3}{Molar\;Mass\;of\;HNO_3}$

Molar mass of $HNO_3$ = 63.01 g/mol

$Moles\;of\;HNO_3=\frac{32.39}{63.01}=0.5140\;mol$

Volume of the solution = 250 mL = 0.250 L

$Molarity=\frac{Moles}{Volume}$

                    =$\frac{0.5140}{0.250} =2.056\;M$