The weight of `Ca(OH)_2` in 100 ml of 0.01 M solution will be: (At. Mass of Ca= 40)
(A) 1.48g
(B) 74g
(C) 14.8g
(D) 0.074g

What is the pH of a solution obtained by mixing 10 mL of 0.1 M HCl and 40 mL 0.2 M H_(2)SO_(4) ?(A) 0.74 (B)7.4 (C)4.68 (D)0.468

In 100ml , 0.1N Na_2CO_3.xH_2O solution. Mass of solute is 1.43g, then value of X is:

A mixture of 100m mol of Ca(OH)_(2) and 2g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH^(-) in resulting solution, respectively, are: (Molar mass of Ca(OH)_(2),Na_(2)SO_(4) and CaSO_(4) are 74, 143 and 136 g mol^(-1) respectively, K_(sp) of Ca(OH)_(2) is 5.5xx10^(-6) )

The volume of 0.1 M Ca (OH)_(2) needed for the neutralization of 40 ml of 0.05 M oxalic acid is

Calcium hydroxide is a strong base. Compute [Ca^(2+)] and [OH^(-)] "for" a solution that is prepared by dissolving 0.60 g of Ca(OH)_(2) in enough water to make a 1500 mL of solution. [Atomic mass : Ca=40, O=16, H=1 ]

Calculate the mass of anhydrous Ca_3(PO_4)_2 present in 100 ml of 0.25 M solution.

Calculate the [OH^(-)] of a solution after 100 mL of 0.1 M MgCl_(2) is added to 100 mL 0.2 M NaOH K_(sp) of Mg(OH)_(2) is 1.2 xx10^(-11) .

0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid is

A mixture of 200 mmol of Ca(OH)_2 and 3g of sodium sulphate was dissolved in water and the volume was made up to 200 mL. The mass of calcium sulphate formed and the concentration of OH^(-) in resulting solution, respectively, are: (Molar mass of Ca(OH)_2 , Na_2SO_4 " and " CaSO_4 are 74, 143 and 136 g mol^(-1) respectively, K_(sp) of Ca (OH)_2 is 5 xx 10^(-6) )

Determine the mass of PbI_(2) that will dissolve in (a) 500mL water (b) 500mL of 0.01M KI solution (c) 500mL of a solution containing 1.33 g Pb(NO_(3))_(2), K_(sp) of PbI = 1.4 xx 10^(-8) .