Some Basic Concepts of Chemistry

This is a Long Answer Type Questions as classified in NCERT Exemplar

The volume of HCl solution is 250 mL and its molarity is 0.76M.

The number of moles of HCl as follows,

Moles of HCl Molarity Volume (in L)

= 0.76M 0.250 L

= 0.19 mol

The molar mass of CaCO₃ is 100 g / gQl and the mass of CaCO₃ is given as 1000 g

The number of moles of CaCO_{3 i}s calculated as

Moles of CaCO₃ =     M a s  / M o l a r   m a s       

=      1000 g / 100 g   /   m o l  = 10 mol

According to the given reaction, 1 mole of CaCO₃ requires 2 moles of HCl. So, the required number of moles of HCl for 10 moles of CaCO₃ is calculated as

Moles of HCl =   2mol of HCl /1mol of CaCO₂   * 10 mol of CaCO₃

=  20 mol

So, the required number of moles of HCl is 20 mol but only 0.19 mol are given. So, HCl is a limiting reagent. So, the amount of calcium chloride formed is depend upon the limiting reagent, that is, the amount of HCl available.

According to the reaction, 2 moles of HCl gives 1 mol of CaCl₂ Sg 2 ., the number of moles of calcium chloride produced by 0.19 mol of HCl as follows,

Moles of CaCl₂ =  1 mol of CaCl₂ / 2 mol  of HCl  x 0.19 mol of HCl

                           = 0.095 mol

The molar mass of calcium chloride is 111 g / mol. So, its mass is calculated as

Mass of CaCl₂ = Moles x  Molar mass

= 0.095 mol x 111 g / mol

= 10.54 g

This is a Long Answer Type Questions as classified in NCERT Exemplar

P₁=1 atm

          P₂= 1/2=0.5 atm

          T₁=273.15 K

          V₂=?

          V₁=?

32 g dioxygen occupies = 22.4 L volume at STP

∴ 1.6 g dioxygen will occupy = 22.4L x 1.6g / 32g = 1.12 L

     V₁=1.12 L

From Boyle's law (as temperature is constant)

p₁V₁=p₂V₂

V₂=p₁V₁p₂

    = 1 atm x 1.12 l/0.6 atm g = 2.24 L

(ii) Number of molecules of dioxygen.

        Ans-  (ii) Number of moles of dioxygen= Mass of dioxygen / Molar mass of dioxygen

                                              nO₂ = 1.6/32 = 0.05 mol
                  1 mol of dioxygen contains = 6.022*10²³ molecules of dioxygen
                  ∴ 0.05 mol of dioxygen = 6.022*10²³*0.05 molecule of O₂

                = 0.3011*10²³ molecules

                =3.011*10²² molecules

1.47. These rules for determining the number of significant figures are stated below:

(1) All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

(2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.

(3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

(4) Zeros at the end or right of a number are significant, provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures. But, if otherwise, the terminal zeros are not significant if there is no decimal point.