Respuesta :
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 [tex]HNO_3 = \frac{70.4}{100} \times 46.008 = 32.39 g[/tex]
[tex]Moles\;of\;HNO_3=\frac{Mass\;of\;HNO_3}{Molar\;Mass\;of\;HNO_3}[/tex]
Molar mass of [tex]HNO_3[/tex] = 63.01 g/mol
[tex]Moles\;of\;HNO_3=\frac{32.39}{63.01}=0.5140\;mol[/tex]
Volume of the solution = 250 mL = 0.250 L
[tex]Molarity=\frac{Moles}{Volume}[/tex]
=[tex]\frac{0.5140}{0.250} =2.056\;M[/tex]
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 [tex]HNO_3[/tex] is dissolved in 100 g of solution.
Density of solution = 1.42 g/ml
Volume of solution = [tex]\frac{mass}{density}=\frac{100g}{1.42g/ml}=70.42ml[/tex]
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{70.4g}{63g/mol}=1.12moles[/tex]
[tex]V_s[/tex] = volume of solution = 70.42 ml
[tex]Molarity=\frac{1.12\times 1000}{70.42}=15.9M[/tex]
According to the neutralization law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HNO_3[/tex] solution = 15.9 M
[tex]V_1[/tex] = volume of stock [tex]HNO_3[/tex] solution = 32.4 ml
[tex]M_2[/tex] = molarity of final [tex]HNO_3[/tex] solution = ?
[tex]V_2[/tex] = volume of final [tex]HNO_3[/tex] solution = 250 ml
[tex]15.9\times 32.4=M\times 250[/tex]
[tex]M=2.06M[/tex]
Therefore, the concentration final solution is 2.06 M.
