Question

determine the molarity of a solution of 4.0 gram / litre concentration and average​

Answer 1

Answer

(a) Molecular mass of NaOH = 40

No. of gram moles in 4 g of NaOH = 40

4 =0.1

Volume of solution in litre = 1000

200 =0.2

Molarity= Volume of solution in litre

No. of moles of solute = 0.2

0.1 =0.5M

(b) Molecular mass of anhydrous Na 2

CO 3=106

No. of gram moles in 5.3 g of anhydrous Na 2

CO 3= 1065.3 =0.05

Volume of solution in litre = 1000

100 =0.1

Molarity = Volume of solution in litre

No. of gram moles of solute= 0.1

0.05=0.5M

(c) Molecular mass of HCl = 36.5

No. of gram moles in 0.365 g of pure HCl = 36.5 0.365

Volume of solution in litre = 1000 50

Molarity= Volume of solutionin litre

No. of gram moles of solute= 36.5

0.365× 50

1000 =0.2M...

$Answer</p><p></p><p>Open in answr app</p><p></p><p>(a) Molecular mass of NaOH = 40</p><p>No. of gram moles in 4 g of NaOH = 404=0.1</p><p>Volume of solution in litre = 1000200=0.2</p><p>Molarity=Volume of solution in litreNo. of moles of solute=0.20.1=0.5M</p><p>(b) Molecular mass of anhydrous Na2CO3=106 </p><p>No. of gram moles in 5.3 g of anhydrous Na2CO3</p><p>=1065.3=0.05</p><p>Volume of solution in litre =1000100=0.1</p><p>Molarity =Volume of solution in litreNo. of gram moles of solute=0.10.05=0.5M</p><p>(c) Molecular mass of HCl = 36.5</p><p>No. of gram moles in 0.365 g of pure HCl =36.50.365</p><p>Volume of solution in litre =100050</p><p>Molarity=Volume of solutionin litreNo. of gram moles of solute</p><p>=36.50.365×501000=0.2M</p><p></p><p>$

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