The Molar Concentration Of A =0.099 .
The Relative Molar Mass Of A = 35.0129 gm
Given,
Mass of HX = 3.5 g
Moles of solution B ([tex]Na_{2} CO_{3}[/tex]) = 0.05 moles
Volume of HX = 26.10 mL
Volume of Solution B = 25 mL
Molecular weight of solution B = 2(atomic weight of Na )+ atomic weight of C + 3(atomic weight of O)
= 2(23) + 12 + 3(16)
=106 gm
Equivalent weight of [tex]Na_{2} CO_{3}[/tex] = molecular weight / 2 = 106 /2 =53 g
Mole = mass / molecular weight
∴0.05 = mass / 106
∴ mass = 5.3 gm [tex]Na_{2} CO_{3}[/tex]
Normality = mass ÷ (equivalent weight × volume of the solution in liter)
= 5.3 ÷( 53 × 0.025)
=4 N
So, by using formula ,
[tex]N_{1} V_{1} =N_{2} V_{2}[/tex]
[tex]N_{1}[/tex] = normality of solution B = 4 N
[tex]V_{1}[/tex] = volume of solution B = 25 mL
[tex]N_{2}[/tex] = normality of solution A = ? N
[tex]V_{2}[/tex] =Volume of solution A = 26.1 mL
∴ 4×25 = [tex]N_{2}[/tex] × 26.1
∴[tex]N_{2}[/tex] = 3.83 N
∴ normality of solution A = 3.83 N
from Formula of the normality we can find the equivalent weight of the A
Normality = mass of HX ÷ (equivalent weight × volume of the solution in liter)
3.83 = 3.5 ÷( equivalent weight × 0.0261)
∴equivalent weight = 35.0129 g
In case of HX the electron transfer is 1 ,so equivalent weight = molecular weight ; which is also termed as relative molar mass in given case.
∴The Relative Molar Mass Of A = 35.0129 g
Molar concentration = mass / molar mass
= 3.5 / 35.0129
= 0.099 mole
∴ The Molar Concentration of A is 0.099 .
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