The molarity of a 0.2 N Na₂CO₅ solution will be :

• 0.05 M

• 0.2 M

• 0.4 M

• 0.1 M

K_sp of CaSO₄.5H₂O is 9 $\times$10^-6, find the volume for 1 g of CaSO₄ (M.wt. = 136).

• 2.45 litres

• 5.1 litres

• 4.52 litres

• 3.2 litres

Which ofthe following sequence contains atomic number of only representative elements?

• 55,12,48,53

• 13,33,54,80

• 3,33,53,87

• 22,33,55,66.

100 cm³ of a given sample of H₂O₂ gives 1000 cm³ of O₂ at S.T.P. The given sample is

• 10% H₂O₂

• 90% H₂O₂

• 10 volume H₂O₂

• 100 volume H₂O₂

325.

Molarity of a given orthophosphoric acid solution is 3 M .It's normality is :

• 9 N

• 0.3 N

• 3 N

• 1 N

Hydroxyl ion concentration of 10^-2 M HCl is :

• 1 x 10¹ mol dm^-3

• 1 x 10^-12 mol dm^-3

• 1 x 10^-1 mol dm^-3

• 1 x 10^-14 mol dm^-3

When 25 g of Na₂SO₄ is dissolved in 10 kg of solution, its concentration will be

• 2.5 ppm

• 25 ppm

• 250 ppm 

• 100 ppm

The volume of 10N and 4N HCl required to make 1 L of 7N HCl are :

• 0.50 L of 10N HCl and 0.50 L of 4N HCl

• 0.60 L of 10N HCl and 0.40 L of 4N HCl

• 0.80 L of 10N HCl and 0.20 L of 4N HCl

• 0.75 L of 10N HCl and 0.25 L of 4N HCl

Stomach acid, a dilute solution of HCl in water,can be neutralized by reaction with sodium hydrogen carbonate,

${\mathrm{NaHCO}}_{3_{\left(\mathrm{aq}\right)}}+ {\mathrm{HCl}}_{\left(\mathrm{aq}\right)}\to {\mathrm{NaCl}}_{\left(\mathrm{aq}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)} + {\mathrm{CO}}_{2\left(\mathrm{g}\right)}$

How many milliliters of 0.125 M NaHCO₂ solution are needed to neutralize 18.0 mL of 0. 100 M HCl?

• 14.4 ml

• 12.0 mL

• 14.0 mL

• 13.2 ml

Assertion : S. l. unit of atomic mass and molecular mass is kilograms.

Reason : Atomic mass is equal to the mass of 6.023x 10²⁴atoms.

• If both assertion and reason are true and reason is the correct explanation of assertion.

• If both assertion and reason are true but reason is not the correct explanation of assertion.

• If assertion is true but reason is false.

• If both assertion and reason are false.