Question

A is a solution containing 3.5g of HX per dm-³. B is a solution containing 0.050moldm-³ of an hydrous sodium trioxocarbonate (IV) of solution 25cm³ portions of solution B required an average of 26.10cm³ of Solution A for complete neutralization. From this results, calculate
1. The Molar Concentration Of A

2. The Relative Molar Mass Of A

Equation For The Reaction 2HX + Na2CO3 ---> 2NaX + H2O + CO2

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Answer 1

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 ($Na_{2} CO_{3}$) = 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  $Na_{2} CO_{3}$ = molecular weight / 2 = 106 /2  =53 g

Mole = mass / molecular weight

∴0.05 = mass / 106

∴ mass = 5.3 gm $Na_{2} CO_{3}$

Normality = mass ÷ (equivalent weight × volume of the solution in liter)

                = 5.3 ÷( 53 × 0.025)

                =4 N

So, by using formula ,

$N_{1} V_{1} =N_{2} V_{2}$

$N_{1}$ = normality of solution B = 4 N

$V_{1}$ = volume of solution B = 25 mL

$N_{2}$ = normality of solution A = ? N

$V_{2}$ =Volume of solution A = 26.1 mL

∴ 4×25 = $N_{2}$ × 26.1

∴$N_{2}$ = 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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