Some Basic Concepts of Chemistry

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% $\ce{O2}$ by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess $\ce{H2S}$ in the presence of conc. HCl ( assuming 100% conversion) is :

An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% sulphur is :

(atomic weight of S = 32 amu)

5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :

The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of $\ce{OH-}$ in aqueous solution is :

The molecular formula of a commercial resin used for exchanging ions in water softening is $\ce{C8H7SO3Na}$ (Mol. wt. 206). What would be the maximum uptake of $\ce{Ca^2+}$ ions by the resin when expressed in mole per gram resin ?

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is : (atomic mass, Ba = 137 amu, Cl = 35.5 amu)

$\ce{A + 2B + 3C <=> AB2C3}$

Reaction of 6.0 g of A, $6.0\times10^{23}$ atoms of B, and 0.036 mol of C yields 4.8 g of compound $\ce{AB2C3}$. If the atomic mass of A and C are 60 and 80 amu, respectively, the atomic mass of B is (Avogadro no. = $6\times10^{23}$) :

If the value of Avogadro number is $6.023\times10^{23}\ \mathrm{mol^{-1}}$ and the value of Boltzmann constant is $1.380\times10^{-23}\ \mathrm{J\,K^{-1}}$, then the number of significant digits in the calculated value of the universal gas constant is

A compound $\ce{H2X}$ with molar weight of 80 g is dissolved in a solvent having density of $0.4\ \mathrm{g\,mL^{-1}}$. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of $\dfrac{M}{10}$ sulphuric acid. The unreacted acid required 20 mL of $\dfrac{M}{10}$ sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is :

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is :

Dissolving 120 g of a compound of (mol. wt. 60) in 1000 g of water gave a solution of density 1.12 g/mL. The molarity of the solution is :

The amount of oxygen in 3.6 moles of water is :

A gaseous compound of nitrogen and hydrogen contains 12.5% (by mass) of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is :

The amount of $\ce{BaSO4}$ formed upon mixing 100 mL of 20.8% $\ce{BaCl2}$ solution with 50 mL of 9.8% $\ce{H2SO4}$ solution will be :

(Ba = 137, Cl = 35.5, S = 32, H = 1 and O = 16)

29.2% (w/w) HCl stock solution has a density of 1.25 g mL⁻¹. The molecular weight of HCl is 36.5 g mol⁻¹. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N $\ce{Na2S2O3}$ was used to reach the end point. The molarity of the household bleach solution is

Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is