NEET Practice Questions (MCQs) with Answers & Solutions

Plan: For the calculation of mass of AgCl precipitated, we find mass of AgNO₃ and NaCl in equal volume with the help of mole concept.

16.9% solution of AgNO₃ means 16.9 g AgNO₃ is present in 100 mL solution.

8.45 g AgNO₃ will present in 50 mL solution

Similarly,

5.8 g NaCl is present in 100 mL solution

2.9 g NaCl is present in 50 mL solution
AgNO₃ + NaCl $\to$AgCl + NaNO₃

Initial mole   8.45/169.8     2.9/58.5       0             0

                     = 0.049          = 0.049  

After reaction     0                 0              0.049     0.049

Mass of AgCl precipitated
= 0.049 x 143.5= 7g

Key Concept Mole is the biggest unit measure of number of molecules/atoms/ions.

._.. 1 mole of water contains molecules = 6.02 x 10²³

.^.. 18 moles of water contain molecules = 18 x 6.02 x 10²³ molecules

Now, 1mole of water = 18 g of water = 6.02 x10²³ and 1.8 g of water contains =6.02 x10²³ molecules

(a) 1.00 molal aqueous solution = 1.0 mole in 1000 g water n_solute = 1; W_solute = 1000 g.

n_solvent = 1000/18 =55.56

X_solute = 1/(1+55.56) = 0.0177

Molarity = number of moles of solute/volume of solution(in mL) x 1000

            = (25.3x1000)/(106x250) = 0.9547 $\approx$0.955 M

Na₂CO₃ in aqueous solution remains dissociated as

        Na₂CO₃$\to$2Na⁺ + ${\mathrm{CO}}_3^{2-}$

           x                  2x               x

Since, the molarity of Na₂CO₃ is 0.955 M, the molarity of ${\mathrm{CO}}_3^{2-}$ is also 0.955 M and that of Na⁺ is

2 x 0.955 = 1.910 M

(b) Number of atoms = number of moles X N_A X atomicity

                          = 0.1 X 6.02 X 10²³ X 3

                          = 1.806 X 10²³ atoms

(a) Number of milliequivalents of HCl = 20 X 0.050 X 1 = 1

Number of milliequivalents of Ba(OH)2 = 2 X 30 X 0.10 = 6

[OH^-] of final solution $=\frac{\mathrm{milliequivalents} \mathrm{of} \mathrm{Ba}(\mathrm{OH}{)}_2 - \mathrm{milliequivalents} \mathrm{of} \mathrm{HCl}}{\mathrm{total} \mathrm{volume}}$

                                 $=\frac{6-1}{50}=0.1 \mathrm{M}$

(c) Key Idea (i) ${\mathrm{H}}_2 +\hspace{0.17em}\frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{H}}_2\mathrm{O}$

(ii) Amount of water produced is decided by limited reactant (i.e., the reactant which is used in small amount)

$$\begin{gathered} {\mathrm{H}}_2 + \hspace{0.17em}\frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{H}}_2\mathrm{O} \\ 1\mathrm{mol} \frac{1}{2}\mathrm{mol} 1\mathrm{mol} \\ \frac{10}{2}\mathrm{mol} \frac{64}{32}\mathrm{mol} ? \\ =5\mathrm{mol} =2\mathrm{mol} \\ \because \frac{1}{2}\mathrm{mol} {\mathrm{O}}_2 \mathrm{gives} = 1\mathrm{mol} {\mathrm{H}}_2\mathrm{O} \\ \therefore 2 \mathrm{mol} {\mathrm{O}}_2 \mathrm{will} \mathrm{give} = 1\times 2\times 2 = 4\mathrm{mol}. \end{gathered}$$

6.023 x 10²³ molecules of water = 1mol =18g

.^.. Mass of one molecule of water = 18/6.023 x10²³ g

 ._.. d=m/V

.^.. V=m/d =18/(6.023x10²³x1) $\approx$3x10^-23 cm³

Key Idea: The reagent which is present in smaller quantity is called the limiting reagent and the moles of product depends on it and number of moles = weight/molecular weight

PbO        +        2HCl  $\to$   PbCl₂  +    H₂O

207.2 +16     2(35.5+1)    207.2+71

 =223.2            =73            =278.2

Mole of PbO = 6.5/223 = 0.029

Mole of HCl = 3.2/36.5 = 0.087

Here, 1 mole of PbO reacts with 2 moles of HCl, thus PbO is the limiting reagent.

._.. 223.2 g PbO gives PbCl₂ = 278.2g

.^.. 6.5g PbO will give PbCl₂

                              =(278.2/223.2) x 6.5g

                              =(278.2 x 6.5)/(223.2 x278.2) mol

                              = 0.029 mol