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"A commerically available sample of sulp...

`"A commerically available sample of sulphuric acid is 15% "H_(2)SO_(4)" by mass" ("density "= 1.10 g mL^(-1)). Calculate :`
Molarity
Normality
Molality of solution.

Text Solution

Verified by Experts

The correct Answer is:
1.68M
3.36M
1.8 M
Calcultion of molality of solution
Mass os `H_(2)SO_(4)` in solution= 15 g
Mass os solvent (water) = 100-5 = 85 g
Molar mass of `H_(2)SO_(4) = 98 " g mol"^(-1)`
`"Molality of solution (m)"=("Mass of" H_(2)SO_(4)//"Molar mass of" H_(2)SO_(4))/("Mass of water in kg")`
`=(15g//(98" g mol"^(-1)))/((85//1000 kg))=1.8 m`
Step II. Calcultation of molarity of solution
`"Density of solution"=1.10" g mol"^(-1)`
`"Volume of 100 g of solution"=("Mass")/("Density")((100g))/((1.10" g mL"^(-1)))=90.9 mL`
`"Molarity of solution (M)"=("Mass of" H_(2)SO_(4)//"Molar mass of" H_(2)SO_(4))/("Volume of solution in litres")`
`((15g)//(98" g mol"^(-1)))/((90.9//1000 L))=1.68 M`.
Step III. Calculatation of normallity of solution
Normality of solution = `"Molarity "xx "Basicity"=(1.68xx2)=3.36 N`.
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A commerically availabe sample of sulphuric acid is 15% H_(2)SO_(4) by weight (density = 1.10 g mL^(-1) ). Calculate (i) molarity (ii) normality and (iii) molality of the solution.

Calculate the molar mass of sulphuric acid (H_(2)SO_(4)) .

Knowledge Check

  • Concentrated aqueous sulphuric acid is 98% H_(2)SO_(4) by weight and has a density of 1.80 gmL^(-1) . Molarity of solution

    A
    1 M
    B
    1.8 M
    C
    18 M
    D
    1.5 M

  • A solution of sulphuric acid in water that is 25% h_(2)SO_(4) by mass has a density of 1.178gmL^(-1) . Which expression gives the molarity of this solution?

    A
    `0.25xx98xx1178`
    B
    `(0.25xx1178)/(98)`
    C
    `(0.25)/(98xx1178)`
    D
    `(1178)/(0.25xx98)`

  • The molarity of H_(2)SO_(4) is 18 M. Its density is 1.8 g mL^(-) . Hence, molality is:

    A
    36
    B
    200
    C
    500
    D
    18

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