NEET Practice Questions (MCQs) with Answers & Solutions

(d) The terms which involves only masses in their formula 

[e.g., molality = mass of solute x 1000 / molar mass of solute x mass of solvent]

are independent of temperature. On the other hand, since, volume changes with temperature, the terms having volume in their formula [e.g., molarity = mass of solute x 1000 / molar mass of solute x volume of solvent],NORMALITY  are dependent on temperatue. 

(d) Dulong Petit's law: atomic mass x sp. heat =6.4

(c) Meq. of H₂SO₄ + Meq.Of SO₃ = Meq. of NaOH

(0.5-a)/49*1000 +a/40*1000 = 26.7 x 0.4

a = 0.103

% of SO₃ = 0.103/0.5*100 = 20.6%

(a)  M₂HPO₄ means valence of metal is one and thus, sulphate of metal is M₂SO₄.

(d) Silica    Water    Clay        Mineral

      45         12         43         Initial %

      a            8        (92-a)      % after heating             

( in dry state ,8% loss of water is there ,so remaining weight is 100 - 8 = 92)

The % ratio of silica and clay remains constant on heating 

i.e., $\frac{45}{43}=\frac{a}{92-a}$

a = 47%

(c) ${\mathrm{\omega }}_{{\mathrm{N}}_2}$ = 1xPx28/RT; ${\mathrm{\omega }}_{\mathrm{CO}}$= 1xPx28/RT; ${\mathrm{\omega }}_{{\mathrm{O}}_2}$=7/8*Px32/RT

(b) M of AgNO₃ = 0.1 x V

M of NaCl = 0.2 x V

M of  N${\mathrm{O}}_3^{-}$ = 0.1 x V and total V= 2V [ Ag NO3 (V) = NO3-(V) , So that total volume becomes 2V]

[ N${\mathrm{O}}_3^{-}$] = 0.1 xV/2V =0.05

(a) For phenolphthalein:

1/2 M eq. of Na₂CO₃ = 2.5x0.1x2=0.5

For methyl orange:

1/2 M eq. of Na₂CO₃ + M eq. of NaHCO₃ = 2.5x0.2x2=1.0

M eq. of NaHCO₃ = 0.5 and M eq. of Na₂CO₃ =1.0

w/84*1000=0.5    w/53*1000=1

w=0.042g in 10 mL       w=0.053g in 10mL

w= 4.2g in 1 litre           w= 5.3 g in 1 litre

(c) Let a g of  Cu be oxidised to give CuO, i.e., $\frac{\left(63.6+16\right)a}{63.6}g$

Thus, final mass = $\left(3.18-1\right)+\frac{\left(63.6+16\right)a}{63.6}$=3.92 

a = 2.94g

Thus, % of Cu left unoxidised =$\frac{\left(3.18-2.94\right)}{3.18}\times 100$  =7.55%